NCERT Exemplar Class 12 Maths Solutions Chapter 12 Linear Programming

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# NCERT Exemplar Class 12 Maths Solutions Chapter 12 Linear Programming

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:19 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 12 Linear Programming - If you often wonder what the best solution to a set of problems would be, then linear programming would help you. It is the process of achieving the best possible outcome of a mathematical model. Explore the detailed concepts in NCERT exemplar Class 12 Maths chapter 12 solutions. Students looking for NCERT Exemplar Class 12 Maths solutions chapter 12 PDF download can use the â€˜Save as webpageâ€™ feature in the browser.
Also, read - NCERT Class 12 Maths Solutions

Question:1

Determine the maximum value of $Z = 11x + 7y$ subject to the constraints: $2x + y \leq 6, x \leq 2, x \geq 0, y \geq 0.$

Given that:
$Z = 11x+7y \\$
It is subject to constraints
$2x + y \leq 6, x \leq 2, x \geq 0, y \geq 0.$
Now let us convert given inequalities into equation.
We obtain following equation
$\\ 2 x+y \leq 6 \\ \Rightarrow 2 x+y=6 \\ x \leq 2 \\ \Rightarrow x=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
The lines that represent 2x+y=6, then further meet the other axes respectively in order to get an answer. The points are to be joined to obtain the line 2x+y=6. It is then further clarified that the equation is satisfied. Then the region that contains the origin is then represented by the set of solutions of the inequation $2x+y \leq 6.\\$
The region represented by $x \leq 2:\\$
The line which is parallel to the Y-axis then meets the X-axis which comes at X=2. Hence, it is clarified that (0,0) satisfies the inequation.
After plotting the equation graphically, we get an answer:

$\\The shaded region OBDE is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.\\ Corner Points are \mathrm{O}(0,0), \mathrm{B}(0,6), \mathrm{D}(2,2) and \mathrm{E}(2,0)$
Coming to the conclusion, when we substitute the values in Z at the corner points, we get the following answer:
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,6) & Z=11(0)+7(6)=0+42=42 \rightarrow \max \\ D(2,2) & Z=11(2)+7(2)=22+14=36 \\ E(2,0) . & Z=11(2)+7(0)=22+0=22 \\ \hline \end{array}$
Therefore, the final answer is the maximum value of Z is 42 at the point (0,6).

Question:2

Maximise $Z = 3x + 4y$, subject to the constraints: $x + y \leq 1, x \geq 0, y \geq 0. \\ \\$

$Z = 3x + 4y$
It is subject to constraints
$x + y \leq 1, x \geq 0, y \geq 0. \\ \\$
Now let us convert the given inequalities into equation.
We obtain the following equation
$\\x+y \leq 1 \\ \Rightarrow x+y=1 \\ x \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
The part represented by $x+y \leq 1:$
One of the lines that is x+y=1 meets the axes (0,1) and (1,0) respectively. Then the lines are joined to obtain the line that is x+y=1. Therefore, it is clear that (0,0) the equation satisfies $x+y \leq 1.$
The region that is represented by $x \geq 0 \and\ y \geq 0$ is first quadrant, and further satisfies these inequations. The graphic plotting is given below:

The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,1) and \mathrm{C}(1,0)$
When we substitute the values in Z, we get the following answer
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x+4 y \\ \hline O(0,0) & Z=3(0)+4(0)=0+0=0 \\ B(0,1) & Z=3(0)+4(1)=0+4=4 \rightarrow \max \\ C(1,0) & Z=3(1)+4(0)=3+0=3 \\ \hline \end{array}$
Hence, the maximum value of Z is 4 at the point (0,1).

Question:3

Maximize the function $Z = 11x + 7y$, subject to the constraints: $x \leq 3, y \leq 2, x \geq 0, y \geq 0.\\$

It is subject to constraints
$x \leq 3, y \leq 2, x \geq 0, y \geq 0.\\$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x \leq 3 \\ \Rightarrow x=3 \\ y \leq 2 \\ \Rightarrow y=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y=0$
The region represented by xâ‰¤3:
The line is parallel to the Y-axis and then meets the X-axis at the point x=3. Then it further gives a clarification that it satisfies the inequation in the problem that is$x \leq 2$. The region then represents the origin and the set of the inequation $x \leq 3.\\$
The region that is represented by $y \leq 2:\\$
The line that is parallel to the x-axis meets the y-axis. The part that contains the region represents the solution set of the other inequation $y \leq 2.\\$
Therefore, the region that represents the $x \geq 0 \ and\ y \geq 0$ is first quadrant and satisfies the inequations. After plotting the graph we get

The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,2), \mathrm{C}(3,2) and \mathrm{D}(3,0)$
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,2) & Z=11(0)+7(2)=0+14=14 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ D(3,0) & Z=11(3)+7(0)=33+0=33 \\ \hline \end{array}$
Therefore, the final answer is that the value of Z is 47 at the point of (3,2).

Question:4

Minimise $Z = 13x -15y$ subject to the constraints: $x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0.\\$

It is given that:
$Z = 13x -15y$
It is subject to constraints
$x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0.\\$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x+y \leq 7 \\ \Rightarrow x+y=7 \\ 2 x-3 y+6 \geq 0 \\ \Rightarrow 2 x-3 y+6=0 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ y=0$
The region which is represented by $x+y \leq 7:\\$
The line in the sum meets the coordinate axes (7,0) and (0,7) respectively. If we join the lines, we will get the other line that is x + y =7. And then it is further clear that (0,0) satisfies the inequation. Then the origin represents the solution for the set of the inequation $x+y \leq 7:\\$
The region represented by $2x - 3y + 6 \geq 0:\\$
The line 2x-3y+6=0 collides with the other axes to coordinate (-3,0) and (0,2) respectively. Then the lines are joined further to obtain the line 2x-3y+6=0. So, the part that contains the origin then represents the other solution set of the inequation $2x - 3y + 6 \geq 0. \\ \\$
Looking at the graph we get,

The shaded region OBCD is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,2), C(3,4) and D(7,0)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=13 x-15 y \\ \hline O(0,0) & Z=13(0)-15(0)=0+0=0 \\ B(0,2) & Z=13(0)-15(2)=0-30=-30 \rightarrow \min \\ C(3,4) & Z=13(3)-15(4)=39-60=-21 \\ D(7,0) & Z=13(7)-15(0)=91-0=91 \end{array}$
So, the final answer of the question is the minimum value of Z is -30 at the point of (0,2).

Question:5

From the question, it is given that
$Z= 3x+4y$
The figure that is given above, from that we can come to a constraint that
$\\ x + 2y \leq 76, 2x +y \leq 104, x \geq 0, y \geq 0\\$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x+2 y \leq 76 \\ \Rightarrow x+2 y=76 \\ 2 x+y \leq 104 \\ \Rightarrow 2 x+y=104 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
The region represented by $x + 2y \leq 76:\\$
We can say that the line $x + 2y=76$ meets the coordinate axes (76,0) and (0,38) respectively. When we join the points to further get the required line we get the line $x + 2y=76$. Then, we can say that it is clear that (0,0) satisfies the inequation $x + 2y \leq 76$ . And then origin is represented by the solution set of inequation$x + 2y \leq 76$
The region represented by $2x +y \leq 104:\\$
The line that has 2x +y=104 then meets the other coordinate axes (52,0) and (0,104) simultaneously. Then we need to join the points to get the result of 2x +y=104. The origin then represents the solution set further of the inequation $2x +y \leq 104:\\$
The first quadrant of the region represented is $x \geq 0 and y \geq 0. \\ \\$
The graph of the equation is given below:

The shaded region ODBA is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), D(0,38), B(44,16) and A(52,0)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x+4 y \\ \hline O(0,0) & Z=3(0)+4(0)=0+0=0 \\ B(0,38) & Z=3(0)+4(38)=0+152=152 \\ C(44,16) & Z=3(44)+4(16)=132+64=196 \rightarrow \max \\ D(52,0) & Z=3(52)+4(0)=156+0=156 \\ \hline \end{array}$
Hence, the maximum value of Z is 196 at the point (44,16)

Question:6

Following is the equation:
Z=5x+7y
The region that is shaded is OABD in the diagram that is being given. The maximum value of the corner point will occur at the feasible point.
When we substitute these values in Z, we get the corner points, we get:
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } \mathrm{Z}=5 \mathrm{x}+7 \mathrm{y} \\ \hline \mathrm{O}(0,0) & \mathrm{Z}=5(0)+7(0)=0+0=0 \\ \mathrm{~A}(7,0) & \mathrm{Z}=5(7)+7(0)=35+0=35 \\ \mathrm{~B}(3,4) & \mathrm{Z}=5(3)+7(4)=15+28=43 \rightarrow \max \\ \mathrm{D}(0,2) & \mathrm{Z}=5(0)+7(2)=0+14=14 \\ \hline \end{array}$
Therefore, the final answer is the maximum value of Z is 43 at the point (3,4)

Question:7

The following is given that:
$Z = 11x + 7y.$
It is subject to constraints
$x+y \leq 5, x+3 y \geq 9, x \geq 0, y \geq 0$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
The region represented by x + y â‰¤ 5:
The line that shows $x+y=5$, then meets the axes that coordinate (5,0) and (0,5) respectively. We need to join the points in order to obtain the line $x+y=5$. Hence, it is clarified that (0,0) then satisfies the inequation $x + y \leq 5$. Therefore, the region that has the origin represents the solution set of $x + y \leq 5$.
The region that is represented by $x +3y \geq 9:\\$
The line that is $x+3y=9$ meets the coordinate axes (9,0) and (0,3) respectively to get the final answer. When we join the points we get the line $x+3y=9$. Therefore, it is then clear that the region doesnâ€™t contain the origin and represents the solution set of the inequation.
The graph is further given below:

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are B(0,3), E(0,5) and C(3,2)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \rightarrow \min \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \\ \hline \end{array}$
Hence, the minimum value of Z is 21 at the point (0,3)

Question:8

Refer to Exercise 7 above. Find the maximum value of Z.

It is given below that:
$Z=11x + 7y$

The figure that is given above, we can see the subject to constraints
$x + y \leq 5, x +3y \geq 9, x \geq 0, y \geq 0\\$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
The region which represents $x + y \leq 5$ is explained below:
The line that is x+y=5, when meets the coordinate axes we get (9,0) and (0,3) respectively. After joining these points, we get the further line that is x+3y=9. The region does not represent the solution set.
The region that represents the first quadrant is $x \geq 0 and y \geq 0. \\$
The graph for the same is given below:

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are B(0,3), E(0,5) and C(3,2)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ \hline \end{array}$
$\text { Hence, the maximum value of } Z \text { is } 47 \text { at the point }(3,2) \text { . }$

Question:9

The feasible region for a LPP is shown in Fig. 12.10. Evaluate $Z = 4x + y$ at each of the corner points of this region. Find the minimum value of Z, if it exists.

It is subject to constraints
$x+2 y \geq 4, x+y \geq 3, x \geq 0, y \geq 0$
Now let us convert the given inequalities into equation
We obtain the following equation
$\\ x+2 y \geq 4 \\ \Rightarrow x+2 y=4 \\ x+y \geq 3 \\ \Rightarrow x+y=3 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0\\ \Rightarrow y=0$
The region representing $x + 2y \geq 4: \\$
The line that is $x + 2y=4$ meets the other coordinate axes (4,0) and (0,2) respectively. Once we join these points we get the result $x + 2y=4$. After joining the lines, it is then clarified that it does not satisfy the inequation $x + 2y \geq 4: \\$. So, then the region that does not contain the origin then represents the set of solutions.
The region that is represented by $x +y \geq 3:\\$
The other line that is $x +y=3$ meets the coordinate axes (3,0) and (0,3) respectively. When we join these points we obtain the line $x +y=3$. It is then cleared that it does not satisfy the inequation. Therefore, the region that does not contain the origin further represents the solution set.
Region represented by $x \geq 0\ and\ y \geq 0$ is first quadrant, since every point in the first quadrant satisfies these inequations
The graph of these equations is given.

The shaded region ABC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are A(0,3), B(2,1), and C(4,0) .
When we see the region, we can see that it is unbounded, and 3 is not the minimum value of Z.
When we decide this issue, we can graph the inequality 4x + y < 3 and then further check that whether the resulting does has the open half no point in common, otherwise Z has no minimum value.
The graph that is shown above, we can see that it is clear that there is no point in common with the region and therefore, Z has a minimum value of 3 at (0, 3).
Henceforth, the minimum value of Z is 3 at the point (0, 3).

Question:10

It is given that:
$Z = x + 2y$
The figure that is given above, from that we can see that the shaded region is feasible and is bounded, so the maximum and the minimum point will occur at the corner point.
\begin{aligned} &\text { Corner Points are } \mathrm{P}\left(\frac{3}{13}, \frac{24}{13}\right), \mathrm{Q}\left(\frac{3}{2}, \frac{15}{4}\right), \mathrm{R}\left(\frac{7}{2}, \frac{3}{4}\right) \text { and } \mathrm{S}\left(\frac{18}{7}, \frac{2}{7}\right)\\ &\text { Now we will substitute these values in } \mathrm{Z} \text { at each of these corner points, we get } \end{aligned}
\begin{aligned} &\begin{array}{|l|l|} \hline \begin{array}{l} \text { Corner } \\ \text { Point } \end{array} & \text { Value of } \mathrm{Z}=\mathrm{x}+2 \mathrm{y} \\ \hline P\left(\frac{3}{13}, \frac{24}{13}\right) & 2=\left(\frac{3}{13}\right)+2\left(\frac{24}{13}\right)=\frac{3}{13}+\frac{48}{13}=\frac{51}{13} \\ Q\left(\frac{3}{2}, \frac{15}{4}\right) & 2=\left(\frac{3}{2}\right)+2\left(\frac{15}{4}\right)=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9 \rightarrow \max \\ R\left(\frac{7}{2}, \frac{3}{4}\right) & 2=\left(\frac{7}{2}\right)+2\left(\frac{3}{4}\right)=\frac{7}{2}+\frac{3}{2}=\frac{10}{2}=5 \\ S\left(\frac{18}{7}, \frac{2}{7}\right) & 2=\left(\frac{18}{7}\right)+2\left(\frac{2}{7}\right)=\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3 \frac{1}{7} \rightarrow \min \\ \hline \end{array}\\ \end{aligned}
$\\ Hence, the maximum value of Z is 9 at the point \left(\frac{3}{2}, \frac{15}{4}\right) .\\ And the minimum value of Z is 3 \frac{1}{7} at the point \left(\frac{18}{7}, \frac{2}{7}\right).$

Question:11

A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.

We can assume that the manufacturer produces x units of type A circuits and y units of type B circuits. We have made the following table from the data that is given to us:
$\begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}$
From the table, we can see that the profit becomes, $Z=50x+60y.\\ \\$
The constraints that we got, i.e. the subject to the constraints,
$20x+10y \leq 200$ [this is resistor constraint]
When we divide it throughout by 10, we get:
$\Rightarrow 2x+y \leq 20 \ldots \ldots \ldots \ldots ..(i)\\ \\$
And $10x+20y \leq 120$ [this is transistor constraint]
After that when we divide it through 10, we get
$\Rightarrow x+3y \leq 15 \ldots \ldots \ldots \ldots ..(iii)\\ \\$
And $x \geq 0, y \geq 0$ [non-negative constraint]
So, the maximum profit is $Z=50x+60y,\\$
subject to $2x+y \leq 20,\\$
$\\ x+2y \leq 12\\ x+3y \leq 15\\ x \geq 0, y \geq 0\\ \\$

Question:12

A firm has to transport 1200 packages using large vans which can carry 200packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.

Let us consider that the firm has X number of large vans and Y number of small vans . We have made the following table:
$\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array}$
Looking at the table, we can see that the cost becomes Z=400x+200y.
Now when we minimize the cost, i.e. minimize $Z=400x+200y$
The subject to constraints are:
$200x+80y \geq1200$
When we divide it by 40, we get:
$\Rightarrow 5x+2y \geq 30 \ldots \ldots \ldots \ldots ..(i)\\ \\$
$And 400 x+200 y \leq 3000$
Now will divide throughout by 200, we get
$\Rightarrow 2 x+y \leq 15$
Also given the number of large vans cannot exceed the number of small vans
$\Rightarrow x \leq y$
And $x \geq 0, y \geq 0$ [non-negative constraint]
So, minimize cost we have to minimize $Z =400 \mathrm{x}+200 \mathrm{y}$ subject to
$\\ 5 x+2 y \geq 30\\ 2 x+y \leq 15 \\ x \leq y\\ x \geq 0, y \geq 0$

Question:13

A company manufactures two types of screws A and B. All the screws have to pass through a threading machine and a slotting machine. A box of Type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours.

On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws.

Formulate this problem as a LPP given that the objective is to maximise profit.

Letâ€™s assume that the company manufactures X boxes of type A screws and Y boxes of type B screws. Look at the following table:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Screws } \\ (\mathrm{x} \text { boxes }) \end{array} & \begin{array}{l} \text { Type B Screws } \\ (\mathrm{y} \text { boxes }) \end{array} & \begin{array}{l} \text { Maximum time } \\ \text { available on } \\ \text { each machine } \\ \text { in a week } \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { threading } \\ \text { machine } \end{array} & 2 & 8 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { slotting } \\ \text { machine } \end{array} & 3 & 2 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \text { Profit } & \text { Rs } 100 &\text { Rs } 170 & \\ \hline \end{array}$
According to the table, we can see that profit becomes $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$
Now, we have to maximize the profit, i.e., maximize $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$
The constraints so obtained, i.e., subject to the constraints,
$2 x+8 y \leq 3600$[time constraints for threading machine]
Now will divide throughout by 2, we get
$\Rightarrow x+4 y \leq 1800$
And $3 x+2 y \leq 3600$ [time constraints for slotting machine]
$\Rightarrow 3 x+2 y \leq 3600 \ldots \ldots \ldots \ldots . .(i i)$
And $x \geq 0,y\geq 0$ [non-negative constraint]
So, to maximize profit we have to maximize $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$ subject to
$\\x+4 y \leq 1800\\ 3 x+2 y \leq 3600\\ x \geq 0, y \geq 0\\$

Question:14

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

Letâ€™s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}$
If we look at the table, we can see that the profit becomes $Z=200x+120y$
If we have to maximize the profit, then maximize $Z=200x+120y$
The constraints so obtained, i.e., subject to the constraints,
The company spends at most Rs 72000 a day.
$\therefore 360 x+120 y \leq 72000$
Divide throughout by 120, we get
$\Rightarrow 3 x+y \leq 600 \ldots...(i)$
Also, company can make at most 300 sweaters.:
$x+y \leq 300 \ldots (ii)$
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100
$i.e., y-x \leq 100$
$\Rightarrow y \leq 100+x \ldots \ldots \ldots$ .....(iii)
And $x \geq 0,y\geq 0$ [non-negative constraint]
So, to maximize profit we have to maximize$Z=200x+120y$ subject to
$\\ 3 x+y \leq 600 \\ \begin{array}{l} x+y \leq 300 \\ y \leq 100+x \\ x \geq 0, y \geq 0 \end{array}$

Question:15

A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hourâ€™s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

If we see the man rides his motorcycle for a distance of X km at a speed of 50km/hr then he will have to spend Rs.2 per km for petrol.
If we assume that the man rides his motorcycle for a distance of Y km at a speed of 80km/hr, then he will need to spend Rs.3 per km on petrol.
Assuming that he has to spend Rs.120 on petrol for a total distance so the constraint becomes,
$2x+3y\leq120.....(i)$
Now also given he has at most one hour's time for total distance to be covered, so the constraint becomes
$\frac{x}{50}+\frac{y}{80} \leq 1 \{as distance =speed x time \}$
Now taking the LCM as 400, we get
$\Rightarrow 8 x+5 y \leq 400$..........(ii)
And $\mathrm{x} \geq 0, \mathrm{y} \geq 0$ [non-negative constraint]
He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$, subject to
$\\2 x+3 y \leq 120 \\8 x+5 y \leq 400 \\x \geq 0, y \geq 0$

Question:16

Refer to Exercise 11. How many circuits of Type A and of Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

If we refer to Exercise 11, we get the following information.
The manufacturer produces X units of type A circuits and y units of type B circuits. We make the following table from the given data:
$\begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}$
Looking at the table, we can see that profit becomes $Z=50x+60y$.
When we maximize the profit, i.e. maximize $Z=50x+60y$.
If we see at the constraints that we have obtained, then the subject to constraints,
$20x+10y \leq 200$ [this is resistor constraint]
Dividing it throughout by 10, we get the answer as:
$2x+y \leq 20 \ldots \ldots \ldots \ldots ..(i)\\ \\$
And $10x+20y \leq 120$ [this is transistor constraint]
Dividing it by 10 throughout we get
$x+2y \leq 12 \ldots \ldots \ldots \ldots ..(ii)\\ \\$
And $10x+30y \leq 150$ [this is capacitor constraint]
Then divide it by 10, we get
$x+3y \leq 15 \ldots \ldots \ldots \ldots ..(iii)\\ \\$
And $x \geq 0, y \geq 0$ [non-negative constraint]
So, when we look at the maximize profit it is $Z=50x+60y$, subject to
$2 x+y \leq 20\\ \\$
$\\ x+2y \leq 12\\ \\ x+3y \leq 15\\ \\ x \geq 0, y \geq 0\\ \\$
When we convert it to equation, we get the following equation
$\\ 2x+y \leq 20\\ \\ 2x+y= 20\\ \\ x+2y \leq 12\\ \\ x+2y= 12\\ \\ x+3y \leq 15\\ \\ x+3y= 15\\ \\ x \geq 0\\ \\ x=0\\ \\ y \geq 0\\ \\ y=0\\ \\$
The region that is represented by $2x+y\leq 20$:
When the line $2x+y=20$ meets the coordinate axes (10,0) and (0,20) respectively. Once after the lines are joined, we obtain the line $2x+y=20$. It is clear that (0,0) satisfies the inequation $2x+y\leq 20$. So the region then represents the set of solutions.
The region represented by $x+2y\leq 12$:
Once the line $x+2y=12$ meets the coordinate axes (12,0) and (0,6) respectively, then the points are joined to get the final result that is $x+2y=12$. It is then clear that it satisfies the inequation and the region does not contain the origin.
The region represented by $x+3y\leq15:$
The line $x+3y=15$ then meets the coordinate axes (15,0) and (0,5) respectively. Once we join these points we obtain the line $x+3y=15$. It is clear that (0,0) satisfies the inequation x+3yâ‰¤ 15. So the region that contains the origin represents the solution set of the inequation $x+3y\leq15:$.
Region represented by xâ‰¥0 and yâ‰¥0 is first quadrant, since every point in the first quadrant satisfies these inequations
The graph of the same is given below:

The shaded region OABCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), A(0,5) , B(6,3), C(9.3,1.3) and D(10,0)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=50 x+60 y \\ \hline 0(0,0) & Z=50(0)+60(0)=0+0=0 \\ A(0,5) & Z=50(0)+60(5)=0+300=300 \\ B(6,3) & Z=50(6)+60(3)-300+180=480 \\ C(9.3,1.3) & Z=50(9.3)+60(1.3)=465+78=543 \rightarrow \max \\ D(10,0) & Z=50(10)+60(0)=500+0=500 \\ \hline \end{array}$
So from the above table the maximum value of Z is at point (9.3,1.3), but as the manufacturer is required to produce two type of circuits, so the parts of resistors, transistors and capacitors cannot be decimals. So we will consider the next maximum number.
Hence, the maximum value of Z is 480 at the point (6,3) i.e., the manufacturer should produce 6 circuits of type A and 3 circuits of type B so as to maximize his profit.

Question:17

Refer to Exercise 12. What will be the minimum cost?

We get the following information referring to the exercise 12:
Let us consider that the firm has X number of large vans and Y number of small vans . We have made the following table:
\begin{aligned} &\text { - }\\ &\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array} \end{aligned}
According to the table, the cost becomes $Z=400x+200y$.
When we have to minimize the cost, i.e., minimize $Z=400x+200y$.
The subject to the constraints are as follows:
$200x+80y \geq 1200\\$
Then divide it by 40, we get
$5x+2y \geq 30 \ldots \ldots \ldots \ldots ..(i)\\$
$And 400x+200y \leq 3000\\$
Then divide it by 200, we get
$2x+y \leq 15 \ldots \ldots \ldots \ldots ..(ii)\\$
The number of the vans that are large exceed the number of small vans
$\\ x \leq y \ldots \ldots \ldots \ldots \ldots ..(iii)\\ \\ And x \geq 0, y \geq 0 [non-negative constraint]\\$
So, when we minimize cost we have to minimize, Z=400x+200y, subject to
$\\ 5x+2y \geq 30\\ \\ 2x+y \leq 15\\ \\ x \leq y\\ \\ x \geq 0, y \geq 0\\ \\$
When we convert the inequalities into equation, we get the following equation
$\\ \\ 5x+2y \geq 30\\ \\ 5x+2y=30\\ \\ 2x+y \leq 15\\ \\ 2x+y=15\\ \\ x \leq y\\ \\ x=y\\ \\ x \geq 0\\ \\ x=0\\ \\ y \geq 0\\ \\ y=0\\ \\$
We can see the region represented by $5x+2y\geq 30$:
The line that is $5x+2y=30$ when it meets the coordinate axes (6,0) and (0,15) respectively we get the desired answer. We will then join these points to obtain the line 5x+2y=30. It is clear that (0,0) does not satisfy the inequation $5x+2y\geq 30$. So the region that does not contain the origin represents the solution set of the inequation $5x+2y\geq 30$.
The region that represents $2x+y\leq15$:
We can see that the line $2x+y=15$ meets the coordinate axes (7.5,0) and (0,15) respectively. We will have to join these points to obtain the line $2x+y=15$. It is clear that (0,0) satisfies the inequation $2x+y\leq15$. So the region that contains the origin represents the solution set of the inequation $2x+y\leq15$.
The region represented by xâ‰¤y:
Then when the line x=y is a line that passes through the origin and doesnâ€™t touch any coordinate axes at any other point except (0,0). When we join the points we get line x=y.
Region represented by xâ‰¥0 and yâ‰¥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The shaded region ABC represents the feasible region is bounded, and minimum value will occur at a corner point of the feasible region.
$\\ Corner Points are \mathrm{A}\left(\frac{30}{7}, \frac{30}{7}\right), \mathrm{B}(0,15) and \mathrm{C}(5,5)$
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=400 \times+200 \mathrm{y} \\ \hline A\left(\frac{30}{7}, \frac{30}{7}\right) & 2=400\left(\frac{30}{7}\right)+200\left(\frac{30}{7}\right) \\ & =\frac{12000}{7}+\frac{6000}{7}=\frac{18000}{7} \\ B(0,15) & =2571.43 \rightarrow \min \\ C(5,5) & Z=400(0)+200(15)=0+3000=3000 \\ & Z=400(5)+200(5)=2000+1000=3000 \\ \hline \end{array}$
So from the above table the minimum value of Z is at point $\left(\frac{30}{7}, \frac{30}{7}\right)$
Hence, the minimum cost of the firm is Rs. 2571.43

Question:18

Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

Refer to the Exercise 13, we get the following data:
Let us assume that the company manufactures x boxes of type A screws and y boxes of type B screws. We make the following table from the given data:

 Type A screws (x boxes) Type B screws (y boxes) Max time available on each machine in a week Time required for screws on threading machine 2 8 60 hrs = 60*60min = 3600min Time required for screws on slotting machine 3 3 60 hrs = 60*60min = 3600min Profit Rs 100 Rs170

When we look at the table, the profit becomes, Z=100x+170y
Thus according to the table, the profit becomes, Z=100x+170y
The constraints that we have obtained that is subject to constraints:
$2x+8y \leq 3600$ [time constraints for threading machine]
Divide it throughout by 2, we get
$x+4y \leq 1800 \ldots \ldots \ldots \ldots ..(i)\\$
And $3x+2y \leq 3600$ [time constraints for slotting machine]
$3x+2y \leq 3600$â€¦â€¦â€¦â€¦..(ii)
And xâ‰¥0, yâ‰¥0 [non-negative constraint]
So, to maximize profit we have to maximize, Z=100x+170y, subject to
$\\ \\ x+4y \leq 1800\\ \\ 3x+2y \leq 3600\\ \\ x \geq 0, y \geq 0\\$
Now let us convert the given inequalities into equation.
We obtain the following equation
$\\ x+4y \leq 1800\\ \\ x+4y=1800\\ \\ 3x+2y \leq 3600\\ \\ 3x+2y=3600\\ \\ x \geq 0\\ \\ x=0\\ \\ y \geq 0\\ \\ y=0\\ \\$
The region that represents x+4yâ‰¤ 1800:
We can say that the line $x+4y=1800$ meets the coordinate axes (1800,0) and (0,450) respectively. When we join these points, we get the desired line that is $x+4y=1800$. We can say that it is clear that it satisfies the inequation and therefore the region that contains the origin further represents the set of solutions of the inequation.
The region represented by $3x+2y\leq3600$:
The line $3x+2y=3600$ further meets the axes (1200,0) and (0,1800) respectively. After joining these points, we get the result that is $3x+2y\leq3600$. It is then clear that it satisfies the inequation $3x+2y\leq3600$. So the region that contains the origin represents the solution set of the inequation $3x+2y\leq3600$.
The graph is given below:

The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,450), C(1080,180) and D(1200,0)
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=100 x+170 y \\ \hline O(0,0) & Z=100(0)+170(0)=0+0=0 \\ B(0,450) & Z=100(0)+170(450)=0+76500=76500 \\ C(1080,180) & Z=100(1080)+170(180)=108000+30600 \\ & Z=138600 \rightarrow \max \\ D(1200,0) & Z=100(1200)+170(0)=120000+0=120000 \\ \hline \end{array}$
So from the above table the maximum value of Z is at point (1080,180) .
Therefore, the final answer is maximum profit to the manufacturer is Rs.
1,38,600

Question:19

Refer to Exercise 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

When we refer to the exercise, we get the following information:
Letâ€™s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}$
Thus according to the table, the profit becomes, Z=200x+120y
If we have to maximize the profit, i.e., maximize Z=200x+120y
The subject to constraints is:
The company spends at most Rs 72000 a day
. $\therefore 360 x+120 y \leq 72000$
Divide throughout by 120, we get
$\Rightarrow 3 x+y \leq 600 \ldots...(i)$
Also, company can make at most 300 sweaters.:$x+y \leq 300 \ldots.. (ii)$
Also, the number of sweaters of type $B$ cannot exceed the number of sweaters of type A by more than 100
i.e., $y-x \leq 100$
$y \leq 100+x \ldots \ldots \ldots(iii)$
And $x \geq 0, y \geq 0$ [non-negative constraint]
\begin{aligned} &\text { So, to maximize profit we have to maximize, } Z=200 x+120 y, \text { subject to }\\ &3 x+y \leq 600\\ &x+y \leq 300\\ &y \leq 100+x\\ &x \geq 0, y \geq 0 \end{aligned}
So, when we convert the inequalities into the equation we get the following answer:
$\\ \mathrm{3x}+\mathrm{y} \leq 600 \\ \Rightarrow 3 \mathrm{x}+\mathrm{y}=600 \\ \mathrm{x}+\mathrm{y} \leq 300 \\ \Rightarrow \mathrm{x}+\mathrm{y}=300 \\ \mathrm{y} \leq 100+\mathrm{x} \\ \Rightarrow \mathrm{y}=100+\mathrm{x} \\ \mathrm{x} \geq 0 \\ \\ \Rightarrow \mathrm{x}=0$
\begin{aligned} &y \geq 0\\ &\Rightarrow y=0\\ &\text { The region represented by } 3 \mathrm{x}+\mathrm{y} \leq 600 \text { : } \end{aligned}
We can say that the line 3x+y=600 meets the coordinate axes (200,0) and (0,600) respectively. Once the points are joined, we get the answer that is 3x+y=600. It is justified that 3x+yâ‰¤ 600. So the region that contain the origin represents the solution set of the inequation 3x+yâ‰¤ 600
The region represented by x+yâ‰¤ 300:
The line when it meets the coordinate axes respectively that is y= 100+x meets (-100,0) and (0,100) respectively and when the lines are joined we get the line y= 100+x. Then it is clarified that it satisfies the inequation. Therefore, the region contains the origin that represents the solution of sets of inequation.
The graph of the same is given:

The shaded region OBCDE is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,100), C(100,200) , D(150,150) and E(200,0)
Now we will substitute these values in Z at each of these corner points, we get
\begin{aligned} &\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=200 x+120 y \\ \hline O(0,0) & Z=200(0)+120(0)=0+0=0 \\ B(0,100) & Z=200(0)+120(100)=0+12000=12000 \\ C(100,200) & Z=200(100)+120(200)=20000+24000=44000 \\ D(150,150) & Z=200(150)+120(150)=30000+18000=48000 \rightarrow \\ E(200,0) & Z=200(200)+120(0)=40000+0=40000 \\ \hline \end{array}\\ &\text { So from the above table the maximum value of } Z \text { is at point }(150,150) \text { . } \end{aligned}
Hence, the final answer is the maximum profit to the manufacturer is Rs. 48,000, for making 150 sweaters each of type A and type B.

Question:20

Refer to Exercise 15. Determine the maximum distance that the man can travel.

When we refer to the Exercise, we get the following information:
If the man rides his motorcycle for a distance of x km at a speed of 50km/hr then he has to spend Rs. 2/km on petrol.
And if he rides his motorcycle for a distance of y km at a speed of 80 km/hr then he has to spend Rs. 3/km on petrol.
He has Rs.120 to spend on petrol for covering the total distance so the constraint becomes
2x+3yâ‰¤120â€¦â€¦â€¦â€¦(i)
Given that he has only one hourâ€™s time to cover the total distance, then the constraint becomes
$\frac{x}{50}+\frac{y}{80} \leq 1$ {as distance speed x time}
Now taking the LCM as 400, we get
$\Rightarrow 8 x+5 y \leq 400 \ldots \ldots \ldots \ldots(ii)$
And $\mathrm{x} \geq 0, \mathrm{y} \geq 0$ [non-negative constraint]
He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$,$subject to $\\2 x+3 y \leq 120 \\8 x+5 y \leq 400 \\x \geq 0, y=0$ When we convert it into equation we get $\\ 2x+3y \leq 120 \Rightarrow 2x+3y=120\\ \\ 8x+5y \leq 400 \Rightarrow 8x+5y=400\\ \\ x \geq 0 \Rightarrow x=0\\ \\ y \geq 0 \Rightarrow y=0\\ \\$ The region that is represented by 2x+3yâ‰¤120: The line that is 2x+3y=120 further meets the coordinate axes (60,0) and (0,40) respectively and then when we will join these points we obtain the line 2x+3y=120. It is clear that (0,0) justifies the inequation 2x+3yâ‰¤120. So the region that contain the origin represents the solution set of the inequation 2x+3yâ‰¤120 The region represented by 8x+5yâ‰¤400: We can see that the line 8x+5y=400 meets the coordinate axes (50,0) and (0,80) respectively. When we join these points we obtain the line 8x+5y=400. Therefore, it is clear that (0,0) satisfies the inequation 8x+5yâ‰¤400. So the region that contain the origin represents the solution set of the inequation 8x+5yâ‰¤400 The first quadrant of the region represented is xâ‰¥0 and yâ‰¥0. The graph is as follows: The shaded region OBCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region. Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,40), \mathrm{C}\left(\frac{300}{7}, \frac{90}{7}\right) and \mathrm{D}(50,0)$ When we substitute the values in Z, we get the following answer: $\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline O(0,0) & Z=0+0=0 \\ B(0,40) & Z=0+40=40 \\ C\left(\frac{300}{7}, \frac{00}{7}\right) & z=\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=54 \frac{2}{7} \rightarrow \max \\ D(50,0) & Z=50+0=50 \\ \hline \end{array}$ So from the above table the maximum value of Z is at point $\left(\frac{300}{7}, \frac{80}{7}\right)$ Hence, the maximum distance the man can travel is $54 \frac{2}{7} \mathrm{~km}$ or 54.3 km Question:21 Maximise $Z = x + y$ subject to $\\ x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0.\\$ Answer: It is given that: $Z = x + y$ And it is also subject to constraints that is given below: $\\ x + 4y \leq 8\\2x + 3y \leq 12\\3x + y \leq 9\\ x \geq 0\\ y \geq 0.\\$ We have to maximize Z, we are subject to the constraints above. We need to convert the inequalities into equation to get the following equation: $\\ \\ x + 4y \leq 8\\ \\ \Rightarrow x + 4y = 8\\ \\ 2x + 3y \leq 12\\ \\ \Rightarrow 2x + 3y = 12\\ \\ 3x + y \leq 9\\ \\ \Rightarrow 3x + y = 9\\ \\ x \geq 0\\ \\ \Rightarrow x=0\\ \\ y \geq 0\\ \\ \Rightarrow y=0\\$ The region that represents $x + 4y \leq 8$ is explained below: The line $x + 4y = 8$ meets the coordinate axes (8,0) and (0,2) respectively. We will join these points to obtain the line $x + 4y = 8$. It is clear that (0,0) satisfies the inequation $x + 4y \leq 8$. So, the region containing the origin represents the solution set of the inequation $x + 4y \leq 8$ The region that represents $2x + 3y \leq 12$: The line that is $2x + 3y = 12$ then meets the coordinate axes respectively to get the answer. When we join the points we obtain the line $2x + 3y = 12$. It is clear that (0,0) satisfies the inequation $2x + 3y \leq 12$. So, the region containing the origin represents the solution set of the inequation$2x + 3y \leq 12$. The region that represents $3x + y \leq 9$: The line meets the coordinate axes that is $3x + y = 9$ meets (3,0) and (0,9) respectively. After joining the lines, we get $3x + y = 9$ and then it is clear that (0,0) satisfies the inequation. The region that contains the origin is represented by the solution set of $3x + y \leq 9$. The graph for the same is given below and also the final answer: $\\Feasible region is ABCD\\ Value of Z at corner points A, B, C and D-$ $\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline \text { A }(2,0) & z=2+0=2 \\ \hline B(2.54,1.36) & Z=2.54+1.36=3.90 \rightarrow \max \\ \hline \text { C }(3,0) & z=3+0=3 \\ \hline \text { D }(0,0) & Z=0+0=0 \\ \hline \end{array}$ $\text { So, value of } Z \text { is maximum at } B(2.54,1.36), \text { the maximum value is } 3.90 .$ Question:22 A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit. Answer: Let us say that the number of bikes per week of model X and Y are x and y respectively. Assuming that model X takes 6 man-hours. So, time taken by x bikes of model X = 6x hours. Assuming that model Y takes 10 man-hours. So, time taken by y bikes of model X = 10y hours. So, the total man-hour that is available per week = 450 So, 6x + 10y â‰¤ 450 3x + 5y â‰¤ 225 The handling and the marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit respectively. So, total handling and marketing cost of x units of model X and y units of model y is 2000x + 1000y The maximum amount that is available for handling and marketing per week is Rs 80000. So, $2000x + 1000y \leq 80000$ $\Rightarrow 2x + y %u2264 80$ Profits per unit for Models X and Y are Rs 1000 and Rs 500, Respectively. Let total profit = Z So, $Z = 1000x + 500y$ Also, as units will be positive numbers so x, y â‰¥ 0 So, we have, $Z = 1000x + 500y$ With constraints, $\\ \\ 3x + 5y \leq 225\\ \\ 2x + y \leq 80\\ \\ x, y \geq 0\\ \\$ In order to maximize Z, that is subject to constraints. We need to convert it into equation: $\\ 3x + 5y \leq 225\\ \\ \Rightarrow 3x + 5y = 225\\ \\ 2x + y \leq 80\\ \\ \Rightarrow 2x + y = 80\\ \\ x \geq 0\\ \\ \Rightarrow x=0\\ \\ y \geq 0\\ \\ \Rightarrow y=0\\$ The graph for the same is given below: ABCD being the feasible region. The Value of Z as well as the final answer is given below. The Value of Z at corner points A,B,C and D : $\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=1000 x+500 y \\ \hline A(0,45) & Z=0+45(500)=22500 \\ \hline B(25,30) & Z=25(1000)+30(500)=4000 \rightarrow \max \\ \hline C(40,0) & Z=40(1000)+0=40000 \rightarrow \max \\ \hline D(0,0) & Z=0+0=0 \\ \hline \end{array}$ So, value of Z is maximum on-line BC, the maximum value is 40000 . So manufacturer must produce 25 number of models X and 30 number of model Y. Question:23 In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are given as below: $\begin{array}{|l|l|l|l|} \hline \text { Tablets } & \text { Iron } & \text { Calcium } & \text { Vitamin } \\ \hline \mathrm{X} & 6 & 3 & 2 \\ \hline \mathrm{Y} & 2 & 3 & 4 \\ \hline \end{array}$ The person needs at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs 2 and Re 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost? Answer: Let us say that the number of tablet X be x and the number of tablet Y be y. Iron content in X and Y tablets is 6 mg and 2 mg respectively. Total iron content from x and y tablets = 6x + 2y Minimum of 18 mg of iron is required. So, we have $\\ 6x + 2y \geq 18\\ \\ 3x + y \geq 9\\$ Similarly, calcium content in X and Y tablets is 3 mg each respectively. So, total calcium content from x and y tablets = 3x + 3y Minimum of 21 mg of calcium is required. So, we have $\\ 6x + 2y \geq 21\\ \\ \Rightarrow x + y \geq 7\\$ Also, vitamin content in X and Y tablet is 2 mg and 4 mg respectively. So, total vitamin content from x and y tablets = 2x + 4y Minimum of 16 mg of vitamin is required. So, we have $\\ 2x + 4y \geq 16\\ \\ \Rightarrow x + 2y \geq 8\\ \\$ Also, as number of tablets should be non-negative so, we have, x, y â‰¥ 0 Cost of each tablet of X and Y is Rs 2 and Re 1 respectively. Let total cost = Z So, Z = 2x + y Finally, we have, Constraints, $\\ 3x + y \geq 9\\ \\ x + y \geq 7\\ \\ x + 2y \geq 8\\ \\ x, y \geq 0\\ \\ Z = 2x + y\\$ We need to minimize Z, subject to the given constraints. Now let us convert the given inequalities into equation. We obtain the following equation $\\ 3x + y \geq 9\\ \\ \Rightarrow 3x + y = 9\\ \\ x + y \geq 7\\ \\ \Rightarrow x + y = 7\\ \\ x + 2y \geq 8\\ \\ \Rightarrow x + 2y = 8\\ \\ x \geq 0\\ \\ \Rightarrow x=0\\ \\ y \geq 0\\ \\ \Rightarrow y=0\\ \\$ The region that is representing 3x + y â‰¥ 9 is the line 3x + y = 9 meets the coordinate axes (3,0) and (0,9) respectively. Once the points are joined, the lines are obtained 3x + y = 9. It does not satisfy the inequation and so the region that contains the origin then represents the solution set of the inequation 3x + y â‰¥ 9. The region that is representing x + y â‰¥ 7 is the line x + y = 7 meets the coordinate axes (7,0) and (0,7) respectively. Once the points are joined, the lines are obtained v It does not satisfy the inequation and so the region that contains the origin then represents the solution set of the inequation x + y â‰¥ 7. The region representing x + 2y â‰¥ 8 is the line x + 2y = 8 meets the coordinate axes (8,0) and (0,4) respectively. We will join these points to obtain the line x + 2y = 8. It is clear that (0,0) does not satisfy the inequation x + 2y â‰¥ 8. So, the region not containing the origin represents the solution set of the inequation x + 2y â‰¥ 8. The regions that represent xâ‰¥0 and yâ‰¥0 is first quadrant, since every point in the first quadrant satisfies these inequations. Given below is the graph: The region towards the right of ABCD is the feasible region. It is unbounded in this case. Value of Z at corner points A,B,C and D : $\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } z=2 x+y \\ \hline A(0,9) & Z=0+9=9 \\ \hline B(1,6) & Z=1(2)+6=8 \rightarrow \text { min } \\ \hline C(6,1) & Z=6(2)+1=13 \\ \hline D(8,0) & Z=8(2)+0=16 \\ \hline \end{array}$ Now, we check if $2 x+y<8$ to check if resulting open half has any point common with feasible region. The region represented by $2 x+y<8$: The line $2x + y=8$meets the coordinate axes (4,0) and (0,8) respectively. We will join these points to obtain the line $2x + y=8$. It is clear that (0,0) satisfies the inequation $2 x+y<8$. So, the region not containing the origin represents the solution set of the inequation $2 x+y<8$. Clearly, $2x + y=8$ intersects feasible region only at B. So, $2 x+y<8$ does not have any point inside feasible region. So, value of Z is minimum at B(1,6), the minimum value is 8 . So, number of tablets that should be taken of type X and Y is 1,6 Respectively. Question:24 A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand. Answer: Taking into consideration that Let number of days for which factory I operate be x and number of days for which factory II operates be y. Number of calculators made by factory I and II of model A are 50 and 40 respectively. Minimum number of calculators of model A required = 6400 So, 50x + 40y â‰¥ 6400 â‡’ 5x + 4y â‰¥ 640 Number of calculators made by factory I and II of model B are 50 and 20 respectively. Minimum number of calculators of model B required = 4000 So, 50x + 20y â‰¥ 4000 â‡’ 5x + 2y â‰¥ 400 Number of calculators made by factory I and II of model C are 30 and 40 respectively. Minimum number of calculators of model C requires = 4800 So, $30x + 40y \geq 4800$ $\Rightarrow 3x + 4y \geq 480$ Operating costs is Rs 12000 and Rs 15000 each day to operate factory I and II respectively. Let Z be total operating cost so we have Z = 12000x + 15000y Also, number of days are non-negative so, x, y â‰¥ 0 So, we have, Constraints, $\\ 5x + 4y \geq 640\\ \\ 5x + 2y \geq 400\\ \\ 3x + 4y \geq 480\\ \\ x, y \geq 0\\ \\ Z = 12000x + 15000y\\$ We need to minimize Z, subject to the given constraints. Now let us convert the given inequalities into equation. We obtain the following equation $\\ 5x + 4y \geq 640\\ \\ \Rightarrow 5x + 4y = 640\\ \\ 5x + 2y \geq 400\\ \\ \Rightarrow 5x + 2y = 400\\ \\ 3x + 4y \geq 480\\ \\ \Rightarrow 3x + 4y = 480\\ \\ x \geq 0\\ \\ \Rightarrow x=0\\ \\ y \geq 0\\ \\ \Rightarrow y=0\\$ The region that represents the 5x + 4y â‰¥ 640. The line 5x + 4y = 640 meets the coordinate axes (128,0) and (0,160) respectively to get the final outcome. When we join these points to obtain the line 5x + 4y = 640. It is then clear that (0,0) does not satisfy the inequation 5x + 4y â‰¥ 640. So, the region not containing the origin represents the solution set of the inequation 5x + 4y â‰¥ 640. The region that represents the 5x + 2y â‰¥ 400: The line 5x + 2y = 400 meets the coordinate axes (80,0) and (0,200) respectively. We will join these points to obtain the line 5x + 2y = 400. It is justified that (0,0) does not satisfy the inequation 5x + 2y â‰¥ 400. So, the region not containing the origin represents the solution set of the inequation 5x + 2y â‰¥ 400. The region represented by 3x + 4y â‰¥ 480: The line that 3x + 4y = 480 when meets the coordinate axes (160,0) and (0,120) respectively. We will then try and join these points to obtain the line 3x + 4y = 480. It is clear that (0,0) does not justify the inequation 3x + 4y â‰¥ 480. So, the region not containing the origin represents the solution set of the inequation 3x + 4y â‰¥ 480. The graph is: The region towards the right of ABCD is the feasible region. It is unbounded in this case. The value of Z at the corner points, is $\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=12000 \times+15000 y \\ \hline A(0,200) & Z=3000000 \\ \hline B(32,120) & Z=218400 \\ \hline C(80,60) & Z=1860000 \rightarrow \text { min } \\ \hline D(160,0) & Z=1920000 \\ \hline \end{array}$ Now, we plot $12000 x+15000 y<1860000$ to check if resulting open half has any point common with feasible region. The region represented by $12000 \mathrm{x}+15000 \mathrm{y}<1860000$ The line $12000 \mathrm{x}+15000 \mathrm{y}=1860000$ meets the coordinate axes (155,0) and (0,124) respectively. We will join these points to obtain the line$$12000 \mathrm{x}+15000 \mathrm{y}=1860000$. It is clear that (0,0) satisfies the inequation $12000 \mathrm{x}+15000 \mathrm{y}<1860000$. So, the region containing the origin represents the solution set of the inequation $12000 \mathrm{x}+15000 \mathrm{y}<1860000$.

Clearly, $12000 \mathrm{x}+15000 \mathrm{y}<1860000$ intersects feasible region only at C
So, value of Z is minimum at C(80,60), the minimum value is 1860000 .
So, number of days factory 1 is required to operate is 80 and number of days factory 2 should operate is 60 to minimize the cost.

Question:25

Maximise and Minimise Z = 3x - 4y

subject to

$\\ x - 2y \leq 0 \\- 3x + y \leq 4 \\x - y \leq 6$

$x,y \geq 0$

We need to maximize and minimize Z, subject to the given constraints.
Now let us convert the given inequalities into equations.
We obtain the following equation
We have constraints,
$\\ x - 2y \leq 0\\ \\ - 3x + y \leq 4\\ \\ x - y \leq 6\\ \\ x, y \geq 0\\ \\ Z = 3x - 4y\\ \\$
We need to maximize and minimize Z, subject to the given constraints.
Now let us convert the given inequalities into equations.
We obtain the following equation
$\\x-2 y \leq 0 \\ \Rightarrow x-2 y=0 \\ -3 x+y \leq 4 \\ \Rightarrow-3 x+y=4 \\ x-y \leq 6$
$\\ \Rightarrow x - y = 6\\ \\ x \geq 0\\ \\ \Rightarrow x=0\\ \\ y \geq 0\\ \\ \Rightarrow y=0\\$
The further explanation of the same is given below:
The region represented by x â€“ 2y â‰¤ 0:
The line x - 2y = 0 meets the coordinate axes at origin and slope of the line is $\frac{1}{2}$. We will construct a line passing through origin and whose slope is $\frac{1}{2}$. As point (1,1) satisfies the inequality. So, the side of the line which contains (1,1) is feasible. Hence, the solution set of the inequation x â€“ 2y â‰¤ 0 is the side which contains (1,1).
The region represented by â€“ 3x + y â‰¤ 4:
The line â€“ 3x + y = 4 meets the coordinate axes $(-\frac{4}{3},0)$ and (0,4) respectively. We will join these points to obtain the line -3x + y = 4. It is clear that (0,0) satisfies the inequation â€“ 3x + y â‰¤ 4. So, the region containing the origin represents the solution set of the inequation â€“ 3x + y â‰¤ 4.
The region represented by x â€“ y â‰¤ 6:
The line x â€“ y = 6 then meets the coordinate axes (6,0) and (0,-6) respectively. We will join these points to obtain the line x â€“ y = 6. It is clear that (0,0) satisfies the inequation x â€“ y â‰¤ 6. So, the region containing the origin represents the solution set of the inequation x â€“ y â‰¤ 6.
The graph for the same is given below:

The feasible region is region between line $-3 x+y=4 \: \: and \: \: x-y=6,$above BC and to the right of y axis as shown.
Feasible region is unbounded.
Corner points are A, B, C
Value of Z at corner points A, B, C and D â€“
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x-4 y \\ \hline A(0,4) & Z=0-(4)(4)=-16 \rightarrow \text { min } \\ \hline B(0,0) & Z=0+0=0 \\ \hline C(12,6) & Z=3(12)-4(6)=12 \rightarrow \max \\ \hline \end{array}$
So, maximum value of Z at corner points is 12 at C and minimum is -16 at A.
So, to check if the solution is correct, we plot 3 x-4 y>12 and 3 x-4 y<-16 for maximum and minimum respectively.
The region represented by 3 x-4 y>12
The line that is 3x â€“ 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. When we join these points, we get the line 3x â€“ 4y > 12. It is clear that (0,0) does not satisfy the inequation 3x â€“ 4y > 12. So, the region not containing the origin represents the solution set of the inequation 3x â€“ 4y > 12.
The region represented by 3x â€“ 4y <-16:
The line 3x â€“ 4y = -16 meets the coordinate axes $(-\frac{16}{3},0)$ and (0,4) respectively. We will join these points to obtain the line 3x â€“ 4y <-16. It is clear that (0,0) does not satisfy the inequation 3x â€“ 4y <-16. So, the region not containing the origin represents the solution set of the inequation 3x â€“ 4y <-16.

Clearly, $3 x-4 y=12$ has no point inside feasible region, but $3 x-4 y=-16$ passes through the feasible region.
Therefore, Z has no minimum value it has only a maximum value which is 12.

Question:26

The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.

Compare the quantity in Column A and Column B.

$\begin{array}{|l|l|} \hline \text { Column A } & \text { Column B } \\ \hline \text { Maximum of Z } & 325 \\ \hline \end{array}$

A. The quantity in column A is greater

B. The quantity in column B is greater

C. The two quantities are equal

D. The relationship cannot be determined on the basis of the information supplied

B)
The quantity in column B is greater
$Z = 4x + 3y$
Corner points- (0, 0), (0, 40), (20, 40), (60, 20), (60, 0)
A feasible region is bounded.
Value of Z at corner points-
At (0, 0), Z = 0
At (0, 40), Z = 120
At (20, 40), Z = 200
At (60, 20), Z = 300
At (60, 0), Z = 240
Clearly, the maximum value of Z is 300, which is less than 325.

Question:27

The feasible solution for a LPP is shown in Fig. 12.12. Let Z = 3x â€“ 4y be the

objective function. Minimum of Z occurs at

A. (0, 0)

B. (0, 8)

C. (5, 0)

D. (4, 10)

B)
Value of Z = 3x â€“ 4y, at corner points are â€“
At (0, 0) = 0
At (0, 8) = -32
At (4, 10) = -28
At (6, 8) = -14
At (6, 5) = -2
At (5, 0) = 15
So, clearly minimum value is at (0, 8)

Question:28

Refer to Exercise 27. Maximum of Z occurs at

A. (5, 0)

B. (6, 5)

C. (6, 8)

D. (4, 10)

Value of Z = 3x â€“ 4y, at corner points are â€“
At (0, 0) = 0
At (0, 8) = -32
At (4, 10) = -28
At (6, 8) = -14
At (6, 5) = -2
At (5, 0) = 15
So, clearly maximum value is at (5, 0)

Question:29

Refer to Exercise 27. (Maximum value of Z + Minimum value of Z) is equal to

A. 13

B. 1

C. â€“ 13

D. â€“ 17

Value of Z = 3x â€“ 4y, at corner points are â€“
At (0, 0) = 0
At (0, 8) = -32
At (4, 10) = -28
At (6, 8) = -14
At (6, 5) = -2
At (5, 0) = 15
So, clearly maximum value is at (5, 0)
Maximum value = 15,
Minimum value = -32
So, maximum + minimum = 15 -32
= -17

Question:30

F = 3x â€“ 4y
Corner points are (0, 0), (0, 4), (12, 6)
Value of F at corner points â€“
At (0, 0), F = 0
At (0, 4), F = -16
At (12, 6), F = 12
Clearly, the maximum value of F = 12.

Question:31

Refer to Exercise 30. Minimum value of F is
A. 0
B. â€“ 16
C. 12
D. does not exist

F = 3x â€“ 4y
Corner points are (0, 0), (0, 4), (12, 6)
Value of F at corner points â€“
At (0, 0), F = 0
At (0, 4), F = -16
At (12, 6), F = 12
Clearly, the minimum value of F = â€“ 16.

Question:32

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function.
The Minimum value of F occurs at

A. (0, 2) only
B. (3, 0) only
C. the midpoint of the line segment joining the points (0, 2) and (3, 0) only
D. any point on the line segment joining the points (0, 2) and (3, 0).

F = 4x + 6y

Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Value of F at corner points â€“
At (0, 2), F = 12
At (3, 0), F = 12
At (6, 0), F = 24
At (6, 8), F = 72
At (0, 5), F = 30
Feasible region â€“
As, feasible region to be bounded so it is a closed polygon.
So, minimum values of F = 12 are at (3, 0) and (0, 2).
Therefore, the minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

Question:33

Refer to Exercise 32, Maximum of F â€“ Minimum of F =
A. 60
B. 48
C. 42
D. 18

F = 4x + 6y
Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Value of F at corner points â€“
At (0, 2), F = 12
At (3, 0), F = 12
At (6, 0), F = 24
At (6, 8), F = 72
At (0, 5), F = 30
Feasible region â€“

Considering, feasible region to be bounded so it is a closed polygon.
Minimum value of F = 12
Maximum value of F = 72
So, Maximum of F â€“ Minimum of F = 60.

Question:34

Given, $Z = px+qy$
Given, minimum occurs at (3, 0) and (1, 1).
For a minimum to occur at two points the value of Z at both points should be the same.
So, value of Z at (3, 0) = value of Z at (1, 1)
â‡’ 3p = p + q
â‡’ 2p = q
$\Rightarrow p=\frac{q}{2}$

So, option B is correct.

Question:35

Fill in the blanks in each of the.
In a LPP, the linear inequalities or restrictions on the variables are called _________.

In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

Question:36

Fill in the blanks in each of the Exercise.
In a LPP, the objective function is always _________

In a LPP, the objective function is always linear.

Question:37

Fill in the blanks in each of the.
If the feasible region for a LPP is _________, then the optimal value of the objective function Z = ax + by may or may not exist.

If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

Question:38

Fill in the blanks in each of the.
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same _________ value.

In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value

Question:39

Fill in the blanks in each of the.
A feasible region of a system of linear inequalities is said to be _________ if it can be enclosed within a circle.

A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.

Question:40

Fill in the blanks in each of the Exercise.
A corner point of a feasible region is a point in the region which is the _________ of two boundary lines.

A corner point of a feasible region is a point in the region which is the intersection of two boundary lines.

Question:41

Fill in the blanks in each of the Exercise.
The feasible region for an LPP is always a _________ polygon.

The feasible region for an LPP is always a convex polygon.

Question:42

State whether the statements in Exercise are True or False.
If the feasible region for a LPP is unbounded, the maximum or minimum of the objective function Z = ax + by may or may not exist.

True
If the feasible region is unbounded then we may or may not have a maximum or minimum of objective function, but if we have a maximum or a minimum value then it must be at one of the corner points only.

Question:43

State whether the statements in Exercise are True or False.
Maximum value of the objective function $Z = ax + by$ in a LPP always occurs at only one corner point of the feasible region.

False.
Maximum value or minimum value can occur at more than one point. In such all the points lie on a line segment and are part of the boundary of the feasible region.

Question:44

State whether the statements in Exercise are True or False.
In a LPP, the minimum value of the objective function $Z = ax + by$ is always 0 if origin is one of the corner points of the feasible region.

False.
Minimum value of objective function can also be negative if the coefficient of x or y is negative. So, it is not necessary that the minimum value of objective function will be zero.

Question:45

State whether the statements in Exercise are True or False.
In a LPP, the maximum value of the objective function $Z = ax + by$ is always finite.

False.
In a LPP, the maximum value of the objective function $Z = ax + by$ may or may not be finite. It depends on the feasible region. If a feasible region is unbounded then we can also have infinite maximum value of objective function.

## More About NCERT Exemplar Solutions for Class 12 Maths Chapter 12

Class 12 Maths NCERT exemplar solutions chapter 12 introduces us to the definition of Linear programming as a mathematical modelling technique in which a function is deemed maximum or minimum when subjected to several constraints. The solution of a linear programming problem targets finding the optimal value of a linear expression.

While some linear programming are quick and does not require a pen and a paper, quite often, the calculations and variables become too complicated but don't worry; NCERT exemplar solutions for Class 12 Maths chapter 12 provided on this page would always give you a quick head-start and aid you in doing well in your 12 board exams and competitive exams.

## Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 12

The topics and subtopics are as follows:

• Linear Programming problems and its Mathematical formulation
• Mathematical formulation of the problem
• Objective function
• Constraints
• Optimisation problems
• Graphical method to solve Linear Programming problems
• Different types of Linear Programming problems
• Manufacturing problems
• Diet problems
• Transportation problems

## What will the students learn in NCERT Exemplar Class 12 Maths Solutions Chapter 12?

• Class 12 Maths NCERT exemplar solutions chapter 12 explores different concepts of Linear Programming. Linear Programming holds immense potential in the field of data analysis and simplifying assumptions.
• NCERT exemplar Class 12 Maths chapter 12 solutions introduces us to the definition of Linear programming as a mathematical modelling technique in which a function is deemed maximum or minimum when subjected to several constraints. The solution of a linear programming problem targets finding the optimal value of a linear expression.
• Through NCERT exemplar Class 12 Maths solutions chapter 12, we would also learn about dozens of real-life applications of Linear Programming in the areas of finance, business planning, industrial engineering, social and physical sciences, production, transportation, and medicine.
• We will learn about several Linear Programming problems, their formulation, and the graphical way to solve them.

## NCERT Exemplar Class 12 Maths Solutions

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Applications of Derivatives Chapter 7 Integrals Chapter 8 Applications of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three dimensional Geometry Chapter 13 Probability

## Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 12

• Some of the broad applications of Linear Programming include areas of food and agriculture, engineering, transportation, manufacturing, and energy. By determining the number of crops and their efficient use, farmers can make the best use of their resources and gather revenue. Engineers often use these problems to help with aerodynamic shape optimisation. The scheduling, travel time, and passengers in the transportation industry are done with Linear Programming. In times of war, linear programming concepts were used to deal with the efficient distributing of resources, keeping in mind the various restrictions such as prices and availability.
• Linear Programming can also be applied to various diet problems, from food aid, national food programs, and dietary guidelines to individual issues after applying certain nutritional constraints, costs constraints, acceptability constraints, and ecological constraints. A student familiar with these applications would have sound concepts and excel in his/her exams.

### NCERT Exemplar Class 12 Solutions

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### Also, check NCERT Solutions for questions given in the book:

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Application of Derivatives Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability

#### Also Check NCERT Books and NCERT Syllabus here

1. How is linear programming an important chapter in maths of Class 12?

These days linear programming is helpful in almost every field and industry from finance to engineering to medicine. Learning this chapter will help in getting a better understanding of mathematical modeling.

2. Are these solutions helpful in entrance exam preparation?

Yes, these NCERT exemplar Class 12 Maths solutions chapter 12 are helpful in the preparation of all types of engineering entrance exams like JEE Main and others.

3. Who prepares these NCERT Class 12 chapter 12 solutions of maths?

These solutions are prepared by our highly experienced experts and teachers, who have years of experience and knowledge regarding NCERT and CBSE pattern and requirement.

4. Are these NCERT solutions exhaustive?

Yes, the NCERT exemplar solutions for Class 12 Maths chapter 12 given here are exhaustive with each step mentioned in complete detail. The solutions are also as per the CBSE guidelines and pattern.

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### Questions related to CBSE Class 12th

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Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects Â and Â we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

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The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

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A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

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If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

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The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

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Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

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The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

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How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

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##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

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##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

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##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

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##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

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##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

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An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

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##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

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Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

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A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

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##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

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A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

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A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

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##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

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##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

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##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

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##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

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##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

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##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

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##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

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##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

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##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

2 Jobs Available
##### Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

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##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available