NCERT Exemplar Class 12 Maths Solutions Chapter 12 Linear Programming

NCERT Exemplar Class 12 Maths Solutions

Linear Programming Exercise 12.3
Page number: 250-257, Total Questions: 45

Question:1

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

Answer:

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
$$\begin{gathered} 2x+y\le 6 \\ \Rightarrow 2x+y=6 \\ x\le 2 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=2 \\ y\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0 \\ y\ge 0 \\ \Rightarrow y=0 \end{gathered}$$
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:
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=11x+7y\\ O(0,0)& Z=11(0)+7(0)=0+0=0\\ B(0,6)& Z=11(0)+7(6)=0+42=42\to max\\ D(2,2)& Z=11(2)+7(2)=22+14=36\\ E(2,0).& Z=11(2)+7(0)=22+0=22\\ \hline\end{array}$
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 $0.$

Answer:

Following is the answer
$Z=3x+4y$
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
$$\begin{gathered} x+y\le 1 \\ \Rightarrow x+y=1 \\ x\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0 \\ y\ge 0 \\ \Rightarrow y=0 \end{gathered}$$
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 $0andy$ \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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=3x+4y\\ O(0,0)& Z=3(0)+4(0)=0+0=0\\ B(0,1)& Z=3(0)+4(1)=0+4=4\to max\\ C(1,0)& Z=3(1)+4(0)=3+0=3\\ \hline\end{array}$
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 $0.$

Answer:

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
$$\begin{gathered} x\le 3 \\ \Rightarrow x=3 \\ y\le 2 \end{gathered}$$
$$\begin{gathered} \Rightarrow y=2 \\ y\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0 \\ y=0 \end{gathered}$$
The region represented by x≤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 $0andy$ \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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=11x+7y\\ O(0,0)& Z=11(0)+7(0)=0+0=0\\ B(0,2)& Z=11(0)+7(2)=0+14=14\\ C(3,2)& Z=11(3)+7(2)=33+14=47\to max\\ D(3,0)& Z=11(3)+7(0)=33+0=33\\ \hline\end{array}$
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 $0.$

Answer:

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
$$\begin{gathered} x+y\le 7 \\ \Rightarrow x+y=7 \\ 2x-3y+6\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x-3y+6=0 \\ y\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0 \\ y\ge 0 \\ y=0 \end{gathered}$$
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
$\overline{)\begin{array}{ll}\text{ Corner Point }& \text{ Value of }Z=13x-15y\\ O(0,0)& Z=13(0)-15(0)=0+0=0\\ B(0,2)& Z=13(0)-15(2)=0-30=-30\to min\\ C(3,4)& Z=13(3)-15(4)=39-60=-21\\ D(7,0)& Z=13(7)-15(0)=91-0=91\end{array}}$
So, the final answer of the question is the minimum value of Z is -30 at the point of (0,2).

Question:5

Determine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for a LPP is shown in Fig.12.7.

Answer:

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
$$\begin{gathered} x+2y\le 76 \\ \Rightarrow x+2y=76 \end{gathered}$$
$2x+y\le 104$
$$\begin{gathered} \Rightarrow 2x+y=104 \\ y\ge 0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=0 \\ y\ge 0 \end{gathered}$$
$\Rightarrow y=0$
The region represented by $x+2y\le 76:$
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 $x+2y\le 76$. And then origin is represented by the solution set of inequation$x+2y\le 76$
The region represented by $2x+y\le 104:$
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 $2x+y\le 104:$
The first quadrant of the region represented is $x\ge 0$ and $y\ge 0.$
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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=3x+4y\\ O(0,0)& Z=3(0)+4(0)=0+0=0\\ B(0,38)& Z=3(0)+4(38)=0+152=152\\ C(44,16)& Z=3(44)+4(16)=132+64=196\to max\\ D(52,0)& Z=3(52)+4(0)=156+0=156\\ \hline\end{array}$
Hence, the maximum value of Z is 196 at the point (44,16)

Question:6

Feasible region (shaded) for a LPP is shown in Fig. 12.8. Maximise $Z=5x+7y.$

Answer:

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:
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }\mathrm{Z}=5\mathrm{x}+7\mathrm{y}\\ \mathrm{O}(0,0)& \mathrm{Z}=5(0)+7(0)=0+0=0\\ \mathrm{A}(7,0)& \mathrm{Z}=5(7)+7(0)=35+0=35\\ \mathrm{B}(3,4)& \mathrm{Z}=5(3)+7(4)=15+28=43\to max\\ \mathrm{D}(0,2)& \mathrm{Z}=5(0)+7(2)=0+14=14\\ \hline\end{array}$
Therefore, the final answer is the maximum value of Z is 43 at the point (3,4)

Question:7

The feasible region for a LPP is shown in Fig. 12.9. Find the minimum value of $Z=11x+7y.$

Answer:

The following is given that:
$Z=11x+7y.$
It is subject to constraints
$x+y\le 5,x+3y\ge 9,x\ge 0,y\ge 0$
Now let us convert the given inequalities into equation
We obtain the following equation
$$\begin{gathered} x+y\le 5 \\ \Rightarrow x+y=5 \\ x+3y\ge 9 \\ \Rightarrow x+3y=9 \\ y\ge 0 \\ \Rightarrow x=0 \\ y\ge 0 \\ \Rightarrow y=0 \end{gathered}$$
The region represented by x + y ≤ 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 doesn’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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=11x+7y\\ B(0,3)& Z=11(0)+7(3)=0+21=21\to min\\ E(0,5)& Z=11(0)+7(5)=0+35=35\\ C(3,2)& Z=11(3)+7(2)=33+14=47\\ \hline\end{array}$
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.

Answer:

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
$$\begin{gathered} x+y\le 5 \\ \Rightarrow x+y=5 \\ x+3y\ge 9 \\ \Rightarrow x+3y=9 \\ y\ge 0 \\ \Rightarrow x=0 \\ y\ge 0 \\ \Rightarrow y=0 \end{gathered}$$
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 $0andy$ \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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=11x+7y\\ B(0,3)& Z=11(0)+7(3)=0+21=21\\ E(0,5)& Z=11(0)+7(5)=0+35=35\\ C(3,2)& Z=11(3)+7(2)=33+14=47\to max\\ \hline\end{array}$
$\text{ Hence, the maximum value of }Z\text{ is }47\text{ at the point }(3,2)\text{ . }$

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.

Answer:

It is subject to constraints
$x+2y\ge 4,x+y\ge 3,x\ge 0,y\ge 0$
Now let us convert the given inequalities into equation
We obtain the following equation
$$\begin{gathered} x+2y\ge 4 \\ \Rightarrow x+2y=4 \\ x+y\ge 3 \\ \Rightarrow x+y=3 \\ y\ge 0 \\ \Rightarrow x=0 \\ y\ge 0 \\ \Rightarrow y=0 \end{gathered}$$
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 $$x\ge 0\$and\$y\ge 0$$ 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

In Fig. 12.11, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of $Z=x+2y$

Answer:

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.
$\begin{array}{rl}& \text{ Corner Points are }\mathrm{P}(\frac{3}{13},\frac{24}{13}),\mathrm{Q}(\frac{3}{2},\frac{15}{4}),\mathrm{R}(\frac{7}{2},\frac{3}{4})\text{ and }\mathrm{S}(\frac{18}{7},\frac{2}{7})\\ & \text{ Now we will substitute these values in }\mathrm{Z}\text{ at each of these corner points, we get }\end{array}$
$\begin{array}{rl}& \begin{array}{|ll|}\hline \begin{array}{c}\text{ Corner }\\ \text{ Point }\end{array}& \text{ Value of }\mathrm{Z}=\mathrm{x}+2\mathrm{y}\\ P(\frac{3}{13},\frac{24}{13})& 2=\left(\frac{3}{13}\right)+2\left(\frac{24}{13}\right)=\frac{3}{13}+\frac{48}{13}=\frac{51}{13}\\ Q(\frac{3}{2},\frac{15}{4})& 2=\left(\frac{3}{2}\right)+2\left(\frac{15}{4}\right)=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9\to max\\ R(\frac{7}{2},\frac{3}{4})& 2=\left(\frac{7}{2}\right)+2\left(\frac{3}{4}\right)=\frac{7}{2}+\frac{3}{2}=\frac{10}{2}=5\\ S(\frac{18}{7},\frac{2}{7})& 2=\left(\frac{18}{7}\right)+2\left(\frac{2}{7}\right)=\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3\frac{1}{7}\to min\\ \hline\end{array}\end{array}$
$$ Hence, the maximum value of Z is 9 at the point $(\frac{3}{2},\frac{15}{4}).$\ And the minimum value of $Z$ is $3\frac{1}{7}$ at the point $(\frac{18}{7},\frac{2}{7})$.$

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.

Answer:

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:
$\begin{array}{|llll|}\hline & \text{ Type A (X) }& \text{ Type B (Y) }& \text{ Maximum Stock }\\ \text{ Resistors }& 20& 10& 200\\ \text{ Transistors }& 10& 12& 120\\ \text{ Capacitors }& 10& 30& 150\\ \text{ Profit }& \text{ Rs. 50 }& \text{ Rs. 60 }& \\ \hline\end{array}$
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,
$20x+10y\le 200$ [this is resistor constraint]
When we divide it throughout by 10, we get:
$\Rightarrow 2x+y\le 20$ ...............(i)
And $10x+20y\le 120$ [this is transistor constraint]
After that when we divide it through 10, we get
$\Rightarrow x+3y\le 15$ ..........(ii)
And $x\ge 0,y\ge 0$ [non-negative constraint]
So, the maximum profit is $Z=50x+60y,$
subject to $2x+y\le 20,$
$$\begin{gathered} x+2y\le 12 \\ x+3y\le 15 \\ x\ge 0,y\ge 0 \end{gathered}$$

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.

Answer:

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

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.

Answer:

Let’s assume that the company manufactures X boxes of type A screws and Y boxes of type B screws. Look at the following table:
$\begin{array}{|llll|}\hline & \begin{array}{c}\text{ Type A Screws }\\ (\mathrm{x}\text{ boxes })\end{array}& \begin{array}{c}\text{ Type B Screws }\\ (\mathrm{y}\text{ boxes })\end{array}& \begin{array}{c}\text{ Maximum time }\\ \text{ available on }\\ \text{ each machine }\\ \text{ in a week }\end{array}\\ \begin{array}{c}\text{ Time }\\ \text{ required for }\\ \text{ screws on }\\ \text{ threading }\\ \text{ machine }\end{array}& 2& 8& \begin{array}{c}60\text{ hours }\\ =60\times 60\min \\ =3600\min \end{array}\\ \begin{array}{c}\text{ Time }\\ \text{ required for }\\ \text{ screws on }\\ \text{ slotting }\\ \text{ machine }\end{array}& 3& 2& \begin{array}{c}60\text{ hours }\\ =60\times 60\min \\ =3600\min \end{array}\\ \text{ Profit }& \text{ Rs }100& \text{ Rs }170& \\ \hline\end{array}$
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,
$2x+8y\le 3600$[time constraints for threading machine]
Now will divide throughout by 2, we get
$\Rightarrow x+4y\le 1800$
And $3x+2y\le 3600$ [time constraints for slotting machine]
$\Rightarrow 3x+2y\le 3600.......(ii)$
And $x\ge 0,y\ge 0$ [non-negative constraint]
So, to maximize profit we have to maximize $\mathrm{Z}=100\mathrm{x}+170\mathrm{y}$ subject to
$x+4y\le 1800,3x+2y\le 3600,x\ge 0,y\ge 0$

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.

Answer:

Let’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:
$\begin{array}{|llll|}\hline & \begin{array}{c}\text{ Type A Sweaters }\\ (\mathrm{X})\end{array}& \begin{array}{c}\text{ Type B Sweaters }\\ (\mathrm{Y})\end{array}& \\ \text{ cost per day }& \text{ Rs. }360& \text{ Rs. }120& \text{ Rs. }72000\\ \begin{array}{c}\text{ Number of }\\ \text{ Sweaters }\end{array}& 1& 1& 300\\ \text{ Profit }& \text{ Rs.200 }& \text{ Rs.120 }& \\ \hline\end{array}$
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.
$\therefore 360x+120y\le 72000$
Divide throughout by 120, we get
$\Rightarrow 3x+y\le 600\dots ...(i)$
Also, company can make at most 300 sweaters.:
$x+y\le 300\dots (ii)$
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100
i.e. $y-x\le 100$
$\Rightarrow y\le 100+x\dots \dots \dots$ .....(iii)
And $x\ge 0,y\ge 0$ [non-negative constraint]
So, to maximize profit we have to maximize $Z=200x+120y$ subject to
$3x+y\le 600$
$x+y\le 300,y\le 100+x,x\ge 0,y\ge 0$

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 hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

Answer:

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,
$2x+3y\le 120.....(i)$
Now also given he has at most one hour's time for total distance to be covered, so the constraint becomes
$\frac{x}{50}+\frac{y}{80}\le 1$ \{as distance =speed x time $\}$
Now taking the LCM as 400, we get
$\Rightarrow 8x+5y\le 400$..........(ii)
And $\mathrm{x}\ge 0,\mathrm{y}\ge 0$ [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
$2x+3y\le 120,8x+5y\le 400,x\ge 0,y\ge 0$

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.

Answer:

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:
$\begin{array}{|llll|}\hline & \text{ Type A (X) }& \text{ Type B (Y) }& \text{ Maximum Stock }\\ \text{ Resistors }& 20& 10& 200\\ \text{ Transistors }& 10& 12& 120\\ \text{ Capacitors }& 10& 30& 150\\ \text{ Profit }& \text{ Rs. 50 }& \text{ Rs. 60 }& \\ \hline\end{array}$
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 $$ \ldots $..(iii)$
And $x$ \geq $0,y$ \geq $0$ [non-negative constraint]
So, when we look at the maximize profit it is $Z=50x+60y$, subject to
$2x+y$ \leq $20$
$x+2y$ \leq $$\begin{gathered} 12 \\ x+3y \end{gathered}$$ \leq $$\begin{gathered} 15 \\ x \end{gathered}$$ \geq $0,y$ \geq $0$
When we convert it to equation, we get the following equation
$2x+y$ \leq $$\begin{gathered} 20 \\ 2x+y=20 \\ x+2y \end{gathered}$$ \leq $$\begin{gathered} 12 \\ x+2y=12 \\ x+3y \end{gathered}$$ \leq $$\begin{gathered} 15 \\ x+3y=15 \\ x \end{gathered}$$ \geq $$\begin{gathered} 0 \\ x=0 \\ y \end{gathered}$$ \geq $$\begin{gathered} 0 \\ y=0 \end{gathered}$$
The region that is represented by $2x+y\le 20$:
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 $2x+y\le 20$. So the region then represents the set of solutions.
The region represented by $x+2y\le 12$:
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 $x+3y\le 15:$
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+3y≤ 15. So the region that contains the origin represents the solution set of the inequation $x+3y\le 15:$.
Region represented by x≥0 and y≥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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=50x+60y\\ 0(0,0)& Z=50(0)+60(0)=0+0=0\\ A(0,5)& Z=50(0)+60(5)=0+300=300\\ B(6,3)& Z=50(6)+60(3)-300+180=480\\ C(9.3,1.3)& Z=50(9.3)+60(1.3)=465+78=543\to max\\ D(10,0)& Z=50(10)+60(0)=500+0=500\\ \hline\end{array}$
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?

Answer:

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:
$\begin{array}{rl}& \text{ - }\\ & \begin{array}{|llll|}\hline & \text{ Large Van (X) }& \text{ Small Van (Y) }& \begin{array}{c}\text{ Maximum/Minim }\\ \text{ um }\end{array}\\ \text{ Packages }& 200& 80& 1200\\ \text{ cost }& \text{ Rs.400 }& \text{ Rs.200 }& \text{ Rs. 3000 }\\ \hline\end{array}\end{array}$
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$\le$ 3000\$Thendivideitby200,weget$2x+y$\le$ 15$\dots$ $\dots$ $\dots$ $\dots$ ..(ii)\$Thenumberofthevansthatarelargeexceedthenumberofsmallvans$\ x$\le$ y$\dots$ $\dots$ $\dots$ $\dots$ $\dots$ ..(iii)\ \ $Andx$ \geq $0,y$ \geq $0[non-negativeconstraint]$
So, when we minimize cost we have to minimize, Z=400x+200y, subject to
$5x+2y$ \geq $$\begin{gathered} 30 \\ 2x+y \end{gathered}$$ \leq $$\begin{gathered} 15 \\ x \end{gathered}$$ \leq $$\begin{gathered} y \\ x \end{gathered}$$ \geq $0,y$ \geq $0$
When we convert the inequalities into equation, we get the following equation
$5x+2y$ \geq $$\begin{gathered} 30 \\ 5x+2y=30 \\ 2x+y \end{gathered}$$ \leq $$\begin{gathered} 15 \\ 2x+y=15 \\ x \end{gathered}$$ \leq $$\begin{gathered} y \\ x=y \\ x \end{gathered}$$ \geq $$\begin{gathered} 0 \\ x=0 \\ y \end{gathered}$$ \geq $$\begin{gathered} 0 \\ y=0 \end{gathered}$$
We can see the region represented by $5x+2y\ge 30$:
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 $5x+2y\ge 30$. So the region that does not contain the origin represents the solution set of the inequation $5x+2y\ge 30$.
The region that represents $2x+y\le 15$:
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 $2x+y\le 15$. So the region that contains the origin represents the solution set of the inequation $2x+y\le 15$.
The region represented by x≤y:
Then when the line x=y is a line that passes through the origin and doesn’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 x≥0 and y≥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.
$$ Corner Points are $\mathrm{A}(\frac{30}{7},\frac{30}{7}),\mathrm{B}(0,15)and\mathrm{C}(5,5)$
Now we will substitute these values in Z at each of these corner points, we get
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=400\times +200\mathrm{y}\\ A(\frac{30}{7},\frac{30}{7})& 2=400\left(\frac{30}{7}\right)+200\left(\frac{30}{7}\right)\\ & =\frac{12000}{7}+\frac{6000}{7}=\frac{18000}{7}\\ B(0,15)& =2571.43\to min\\ C(5,5)& Z=400(0)+200(15)=0+3000=3000\\ & Z=400(5)+200(5)=2000+1000=3000\\ \hline\end{array}$
So from the above table the minimum value of Z is at point $(\frac{30}{7},\frac{30}{7})$$
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.

Answer:

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:

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 $$ \ldots $..(i)$
And $3x+2y$ \leq $3600$ [time constraints for slotting machine]
$3x+2y$ \leq $3600$…………..(ii)
And x≥0, y≥0 [non-negative constraint]
So, to maximize profit we have to maximize, Z=100x+170y, subject to
$x+4y$ \leq $$\begin{gathered} 1800 \\ 3x+2y \end{gathered}$$ \leq $$\begin{gathered} 3600 \\ x \end{gathered}$$ \geq $0,y$ \geq $0$
Now let us convert the given inequalities into equation.
We obtain the following equation
$x+4y$ \leq $$\begin{gathered} 1800 \\ x+4y=1800 \\ 3x+2y \end{gathered}$$ \leq $$\begin{gathered} 3600 \\ 3x+2y=3600 \\ x \end{gathered}$$ \geq $$\begin{gathered} 0 \\ x=0 \\ y \end{gathered}$$ \geq $$\begin{gathered} 0 \\ y=0 \end{gathered}$$
The region that represents x+4y≤ 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 $3x+2y\le 3600$:
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 $3x+2y\le 3600$. It is then clear that it satisfies the inequation $3x+2y\le 3600$. So the region that contains the origin represents the solution set of the inequation $3x+2y\le 3600$.
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
$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=100x+170y\\ O(0,0)& Z=100(0)+170(0)=0+0=0\\ B(0,450)& Z=100(0)+170(450)=0+76500=76500\\ C(1080,180)& Z=100(1080)+170(180)=108000+30600\\ & Z=138600\to max\\ D(1200,0)& Z=100(1200)+170(0)=120000+0=120000\\ \hline\end{array}$
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?

Answer:

When we refer to the exercise, we get the following information:
Let’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:
$\begin{array}{|llll|}\hline & \begin{array}{c}\text{ Type A Sweaters }\\ (\mathrm{X})\end{array}& \begin{array}{c}\text{ Type B Sweaters }\\ (\mathrm{Y})\end{array}& \\ \text{ cost per day }& \text{ Rs. }360& \text{ Rs. }120& \text{ Rs. }72000\\ \begin{array}{c}\text{ Number of }\\ \text{ Sweaters }\end{array}& 1& 1& 300\\ \text{ Profit }& \text{ Rs.200 }& \text{ Rs.120 }& \\ \hline\end{array}$
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
. $$\therefore 360x+120y\le 72000$$
Divide throughout by 120, we get
$$\Rightarrow 3x+y\le 600\dots \$...(i)\$Also,companycanmakeatmost300sweaters.:$$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., $$y-x\le 100$$
$$y\le 100+x\dots \dots \dots \$(iii)\$And$$x \geq 0, y \geq 0$$ [non-negative constraint]
$\begin{array}{rl}& \text{ So, to maximize profit we have to maximize, }Z=200x+120y,\text{ subject to }\\ & 3x+y\le 600\\ & x+y\le 300\\ & y\le 100+x\\ & x\ge 0,y\ge 0\end{array}$
So, when we convert the inequalities into the equation we get the following answer:
$$\begin{gathered} 3\mathrm{x}+\mathrm{y}\le 600 \\ \Rightarrow 3\mathrm{x}+\mathrm{y}=600 \\ \mathrm{x}+\mathrm{y}\le 300 \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{x}+\mathrm{y}=300 \\ \mathrm{y}\le 100+\mathrm{x} \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{y}=100+\mathrm{x} \\ \mathrm{x}\ge 0 \\ \Rightarrow \mathrm{x}=0 \end{gathered}$$
$\begin{array}{rl}& y\ge 0\\ & \Rightarrow y=0\\ & \text{ The region represented by }3\mathrm{x}+\mathrm{y}\le 600\text{ : }\end{array}$
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+y≤ 600. So the region that contain the origin represents the solution set of the inequation 3x+y≤ 600
The region represented by x+y≤ 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
$\begin{array}{rl}& \begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=200x+120y\\ O(0,0)& Z=200(0)+120(0)=0+0=0\\ B(0,100)& Z=200(0)+120(100)=0+12000=12000\\ C(100,200)& Z=200(100)+120(200)=20000+24000=44000\\ D(150,150)& Z=200(150)+120(150)=30000+18000=48000\to \\ E(200,0)& Z=200(200)+120(0)=40000+0=40000\\ \hline\end{array}\\ & \text{ So from the above table the maximum value of }Z\text{ is at point }(150,150)\text{ . }\end{array}$
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.

Answer:

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+3y≤120…………(i)
Given that he has only one hour’s time to cover the total distance, then the constraint becomes
$\frac{x}{50}+\frac{y}{80}\le 1$ {as distance speed x time}
Now taking the LCM as 400, we get
$$\Rightarrow 8 x+5 y \leq 400 \ldots \ldots \ldots \ldots$(ii)$
And $\mathrm{x}\ge 0,\mathrm{y}\ge 0$ [non-negative constraint]
He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$, subject to
$2x+3y\le 120$ \$8x+5y\le 400$ \$x\ge 0,y=0$
When we convert it into equation we get
$2x+3y$ \leq $120$ \Rightarrow $$\begin{gathered} 2x+3y=120 \\ 8x+5y \end{gathered}$$ \leq $400$ \Rightarrow $$\begin{gathered} 8x+5y=400 \\ x \end{gathered}$$ \geq $0$ \Rightarrow $$\begin{gathered} x=0 \\ y \end{gathered}$$ \geq $0$ \Rightarrow $y=0$
The region that is represented by 2x+3y≤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+3y≤120. So the region that contain the origin represents the solution set of the inequation 2x+3y≤120
The region represented by 8x+5y≤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+5y≤400. So the region that contain the origin represents the solution set of the inequation 8x+5y≤400
The first quadrant of the region represented is x≥0 and y≥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 $\mathrm{O}(0,0),\mathrm{B}(0,40),\mathrm{C}(\frac{300}{7},\frac{90}{7})$ and $\mathrm{D}(50,0)$$WhenwesubstitutethevaluesinZ,wegetthefollowinganswer:$$$\begin{array}{|ll|}\hline \text{ Corner Point }& \text{ Value of }Z=x+y\\ O(0,0)& Z=0+0=0\\ B(0,40)& Z=0+40=40\\ C(\frac{300}{7},\frac{00}{7})& z=\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=54\frac{2}{7}\to max\\ D(50,0)& Z=50+0=50\\ \hline\end{array}$$$SofromtheabovetablethemaximumvalueofZisatpoint$\left(\frac{300}{7}, \frac{80}{7}\right)$Hence,themaximumdistancethemancantravelis$54 \frac{2}{7} \mathrm{~km}$ or 54.3 km