NCERT Exemplar Class 12 Maths Solutions Chapter 12 Linear Programming

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

Question:7

The 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
$x+y \leq 5, x+3 y \geq 9, x \geq 0, y \geq 0$
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 ≤ 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
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \rightarrow \min \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \\ \hline \end{array}$
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.

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
$\\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0$
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.

Answer:

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

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

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:
$\begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}$
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:
$\Rightarrow 2x+y \leq 20$ ...............(i)
And $10x+20y \leq 120$ [this is transistor constraint]
After that when we divide it through 10, we get
$\Rightarrow x+3y \leq 15$ ..........(ii)
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.

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:
$\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array}$
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:
$\Rightarrow 5x+2y \geq 30$ .........(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
$5 x+2 y \geq 30,\ 2x+y \leq 15, \ x \leq y,\ x \geq 0, y \geq 0$

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.

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:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Screws } \\ (\mathrm{x} \text { boxes }) \end{array} & \begin{array}{l} \text { Type B Screws } \\ (\mathrm{y} \text { boxes }) \end{array} & \begin{array}{l} \text { Maximum time } \\ \text { available on } \\ \text { each machine } \\ \text { in a week } \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { threading } \\ \text { machine } \end{array} & 2 & 8 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { slotting } \\ \text { machine } \end{array} & 3 & 2 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \text { Profit } & \text { Rs } 100 &\text { Rs } 170 & \\ \hline \end{array}$
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]
$\Rightarrow 3 x+2 y \leq 3600 .......(ii)$
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.

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:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}$
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
$3x+y \leq 600$
$x+y \leq 300,\ y \leq 100+x,\ x \geq 0,\ y \geq 0$

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,
$2x+3y\leq120.....(i)$
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.

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:
$\begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}$
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+3y≤ 15. So the region that contains the origin represents the solution set of the inequation $x+3y\leq15:$.
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
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=50 x+60 y \\ \hline 0(0,0) & Z=50(0)+60(0)=0+0=0 \\ A(0,5) & Z=50(0)+60(5)=0+300=300 \\ B(6,3) & Z=50(6)+60(3)-300+180=480 \\ C(9.3,1.3) & Z=50(9.3)+60(1.3)=465+78=543 \rightarrow \max \\ D(10,0) & Z=50(10)+60(0)=500+0=500 \\ \hline \end{array}$
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?

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:
$\begin{aligned} &\text { - }\\ &\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array} \end{aligned}$
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 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}\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.

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 $ 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+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 $3x+2y\leq3600$:
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?

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:
$\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}$
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+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
$\begin{aligned} &\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=200 x+120 y \\ \hline O(0,0) & Z=200(0)+120(0)=0+0=0 \\ B(0,100) & Z=200(0)+120(100)=0+12000=12000 \\ C(100,200) & Z=200(100)+120(200)=20000+24000=44000 \\ D(150,150) & Z=200(150)+120(150)=30000+18000=48000 \rightarrow \\ E(200,0) & Z=200(200)+120(0)=40000+0=40000 \\ \hline \end{array}\\ &\text { So from the above table the maximum value of } Z \text { is at point }(150,150) \text { . } \end{aligned}$
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.

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
$\frac{x}{50}+\frac{y}{80} \leq 1$ {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+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 $\mathrm{O}(0,0), \mathrm{B}(0,40), \mathrm{C}\left(\frac{300}{7}, \frac{90}{7}\right)$ 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, the time taken by x bikes of model X = 6x hours.
Assuming that model Y takes 10 man-hours.
So, the time taken by y bikes of model X = 10y hours.
So, the total man-hour that is available per week = 450
So, 6x + 10y ≤ 450
3x + 5y ≤ 225
The handling and the marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit, respectively.
So, the 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, $2000x + 1000y \leq 80000$
$\Rightarrow 2x + y ? 80$
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 ≥ 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, which is subject to constraints.
We need to convert it into an 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 is x and the number of tablet Y be y.
The iron content in X and Y tablets is 6 mg and 2 mg, respectively.
Total iron content from x and y tablets = 6x + 2y
A minimum of 18 mg of iron is required. So, we have
$\\ 6x + 2y $ \geq $ 18\\ \\ 3x + y $ \geq $ 9\\$
Similarly, the calcium content in X and Y tablets is 3 mg each, respectively.
So, total calcium content from x and y tablets = 3x + 3y
A minimum of 21 mg of calcium is required. So, we have
$\\ 6x + 2y $ \geq $ 21\\ \\ $ \Rightarrow $ x + y $ \geq $ 7\\$
Also, the vitamin content in X and Y tablets is 2 mg and 4 mg, respectively.
So, total vitamin content from x and y tablets = 2x + 4y
A minimum of 16 mg of vitamin is required. So, we have
$\\ 2x + 4y $ \geq $ 16\\ \\ $ \Rightarrow $ x + 2y $ \geq $ 8\\ \\$
Also, as the number of tablets should be non-negative so, we have
x, y ≥ 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 represented 3x + y ≥ 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 ≥ 9.
The region that is represented x + y ≥ 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 ≥ 7.
The region representing x + 2y ≥ 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 ≥ 8. So, the region not containing the origin represents the solution set of the inequation x + 2y ≥ 8.
The regions that represent x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequalities.
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 the feasible region.
So, value of Z is minimum at B(1,6), the minimum value is 8.
So, the 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:

Let the number of days for which factory I operates be x and the number of days for which factory II operates be y.
The number of calculators made by factory I and II of model A is 50 and 40, respectively.
Minimum number of calculators of model A required = 6400
So, 50x + 40y ≥ 6400
⇒ 5x + 4y ≥ 640
The number of calculators made by factory I and II of model B is 50 and 20, respectively.
Minimum number of calculators of model B required = 4000
So, 50x + 20y ≥ 4000
⇒ 5x + 2y ≥ 400
The number of calculators made by factory I and II of model C are 30 and 40, respectively.
The minimum number of calculators of model C required = 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, the number of days is non-negative so x, y ≥ 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 an 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 ≥ 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 inequality 5x + 4y ≥ 640. So, the region not containing the origin represents the solution set of the inequality 5x + 4y ≥ 640.
The region that represents the 5x + 2y ≥ 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 inequality 5x + 2y ≥ 400. So, the region not containing the origin represents the solution set of the inequality 5x + 2y ≥ 400.
The region represented by 3x + 4y ≥ 480:
The line that 3x + 4y = 480 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 ≥ 480. So, the region not containing the origin represents the solution set of the inequality 3x + 4y ≥ 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 the number of days factory 2 should operate is 60 to minimize the cost.

Question:25

Maximise and Minimise Z = 3x - 4y

subject to

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

$x,y \geq 0$

Answer:

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 – 2y ≤ 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 inequality x – 2y ≤ 0 is the side which contains (1,1).
The region represented by – 3x + y ≤ 4:
The line – 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 inequality – 3x + y ≤ 4. So, the region containing the origin represents the solution set of the inequality – 3x + y ≤ 4.
The region represented by x – y ≤ 6:
The line x – y = 6 then meets the coordinate axes (6,0) and (0,-6) respectively. We will join these points to obtain the line x – y = 6. It is clear that (0,0) satisfies the inequality x–y ≤ 6. So, the region containing the origin represents the solution set of the inequality x – y ≤ 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 –
$\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x-4 y \\ \hline A(0,4) & Z=0-(4)(4)=-16 \rightarrow \text { min } \\ \hline B(0,0) & Z=0+0=0 \\ \hline C(12,6) & Z=3(12)-4(6)=12 \rightarrow \max \\ \hline \end{array}$
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 – 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. When we join these points, we get the line 3x – 4y > 12. It is clear that (0,0) does not satisfy the inequality 3x – 4y > 12. So, the region not containing the origin represents the solution set of the inequality 3x – 4y > 12.
The region represented by 3x – 4y <-16:
The line 3x – 4y = -16 meets the coordinate axes $(-\frac{16}{3},0)$ and (0,4) respectively. We will join these points to obtain the line 3x – 4y <-16. It is clear that (0,0) does not satisfy the inequality 3x – 4y <-16. So, the region not containing the origin represents the solution set of the inequality 3x – 4y <-16.
We get the following answer:

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

Answer:

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 – 4y be the

objective function. Minimum of Z occurs at

A. (0, 0)

B. (0, 8)

C. (5, 0)

D. (4, 10)

Answer:

B)
Answer
Value of Z = 3x – 4y, at corner points are –
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)

Answer:

Correct Answer A
Answer
Value of Z = 3x – 4y, at corner points are –
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. – 13

D. – 17

Answer:

Correct Answer D)
Answer:
Value of Z = 3x – 4y, at corner points are –
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

The feasible region for an LPP is shown in the Fig. 12.13. Let F = 3x – 4y be the objective function. Maximum value of F is.

A. 0
B. 8
C. 12
D. – 18

Answer:

Correct Answer C)
Answer:
F = 3x – 4y
Corner points are (0, 0), (0, 4), (12, 6)
Value of F at corner points –
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. – 16
C. 12
D. does not exist

Answer:

Correct Answer B)
Answer:
F = 3x – 4y
Corner points are (0, 0), (0, 4), (12, 6)
Value of F at corner points –
At (0, 0), F = 0
At (0, 4), F = -16
At (12, 6), F = 12
Clearly, the minimum value of F = – 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).

Answer:

Correct Answer D)
Answer:
F = 4x + 6y

Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Value of F at corner points –
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 –
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 – Minimum of F =
A. 60
B. 48
C. 42
D. 18

Answer:

Correct Answer A)
Answer:
F = 4x + 6y
Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Value of F at corner points –
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 –

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 – Minimum of F = 60.

Question:34

Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let $Z = px+qy$ , where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is
A. p = 2q
B. p = q/2
C. p = 3q
D. p = q

Answer:

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)
⇒ 3p = p + q
⇒ 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 _________.

Answer:

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 _________

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.