Solution

Given that: $Z=11 x+7 y$ and the constraints $2 x+y \leq 6, x \leq 2$,

$x \geq 0, y \geq 0$

Let $2 x+y=6$

$\begin{array}{|l|l|l|} \hline x & 0 & 3 \\ \hline y & 6 & 0 \\ \hline \end{array}$

The shaded area OABC is the feasible region determined by the constraints

$2 x+y \leq 6, x \leq 2, x \geq 0, y \geq 0$

The feasible region is bounded.

So, maximum value will occur at a corner point of the feasible region.

Corner points are $(0,0),(2,0),(2,2)$ and $(0,6)$.

Now, evaluating the value of $Z$, we get

Hence, the maximum value of $Z$ is 42 at $(0,6)$.

Solution

Given that: $Z=3 x+4 y$ and the constraints $x+y \leq 1, x \geq 0, y \geq 0$

Let

$ x+y=1 $

$\begin{array}{|l|l|l|} \hline x & 1 & 0 \\ \hline y & 0 & 1 \\ \hline \end{array}$

The shaded area $OAB$ is the feasible region determined by $x+y \leq 1, x \geq 0, y \geq 0$.

The feasible region is bounded. So, maximum value will occur at the corner points $O(0,0)$, $A(1,0), B(0,1)$.

Now, evaluating the value of $Z$, we get

Hence, the maximum value of $Z$ is 4 at $(0,1)$.

Solution

The shaded region is the feasible region determined by the constraints $x \leq 3, y \leq 2$, $x \geq 0, y \geq 0$.

The feasible region is bounded with four corners $O(0,0), A(3,0), B(3,2)$ and

$C(0,2)$.

So, the maximum value can occur at any corner.

Let us evaluate the value of $Z$.

Hence, the maximum value of the function $Z$ is 47 at $(3,2)$.

Solution

Given that: $Z=13 x-15 y$ and the constraints $x+y \leq 7,2 x-3 y+6 \geq 0, x \geq 0, y \geq 0$

Let $x+y=7$

$\begin{array}{|l|l|l|} \hline x & 3 & 4 \\ \hline y & 4 & 3 \\ \hline \end{array}$

Let $2 x-3 y+6=0$

$\begin{array}{|l|l|l|} \hline x & 1 & -3 \\ \hline y & 2 & 0 \\ \hline \end{array}$

The shaded region is the feasible region determined

The feasible region is bounded with four corners

$O(0,0), A(7,0), B(3,4), C(0,2)$

So, the maximum value can occur at any corner.

Let us evaluate the value of $Z$.

Hence, the minimum value of $Z$ is -30 at $(0,2)$.

Solution

As shown in the figure, OAED is the feasible region.

At A, $y=0 \therefore 2 x+y=104$

$\Rightarrow x=52$

Which gives corner point $A=(52,0)$

At D, $x=0 \therefore x+2 y=76 \Rightarrow y=38$

Which gives corner point $D=(0,38)$

Now solving the given equations, we get

$ \begin{aligned} x+2 y=76 \\ 2 x+y=104 \\ 2 x+4 y=152 \\ 2 x+y=104 \\ \hline (-) \quad (-) \quad (-) \\ \qquad 3 y=48 \Rightarrow y=16 \end{aligned} $

$x+2(16) =76$

$\Rightarrow \quad x =76-32=44$

So, the corner point $E=(44,16)$

Evaluating the maximum value of $Z$, we get

Hence, the maximum value of $Z$ is 196 at $(44,16)$.

Solution

$OABC$ is the feasible region whose corner points are $O(0,0), A(7,0), B(3,4)$ and $C(0,2)$

Evaluating the value of $Z$, we get

Hence, the maximum value of $Z$ is 43 at $(3,4)$.

Solution

As per the given figure, ABCA is the feasible region. Corner points $C(0,3), B(0,5)$ and for $A$, we have to solve equations

$ \text{ and } \quad \begin{aligned} x+3 y & =9 \\ x+y & =5 \end{aligned} $

Which gives $x=3, y=2$

i.e., $A(3,2)$

Evaluating the value of $Z$, we get

Hence, the minimum value of $Z$ is 21 at $(0,3)$.

Solution

As per the evaluating table for the value of $Z$, it is clear that the maximum value of $Z$ is 47 at $(3,2)$.

Solution

As per the given figure, $A B C$ is the feasible region which is open unbounded.

Here, we have

$$ \begin{align*} x+y & =3 \tag{i}\\ \text{ and } x+2 y & =4 \tag{ii}\\ Z & =4 x+y \end{align*} $$

Solving eq. (i) and (ii), we get

$x=2$ and $y=1$

So, the corner points are $A(4,0), B(2,1)$ and $C(0,3)$

Let us evaluate the value of $Z$

Now, the minimum value of $Z$ is 3 at $(0,3)$ but since, the feasible region is open bounded so it may or may not be the minimum value of $Z$.

Therefore, to face such situation, we draw a graph of $4 x+y<3$ and check whether the resulting open half plane has no point in common with feasible region. Otherwise $Z$ will have no minimum value. From the graph, we conclude that there is no common point with the feasible region.

Hence, $Z$ has the minimum value 3 at $(0,3)$.

Solution

Here, corner points are given as follows:

$ R(\frac{7}{2}, \frac{3}{4}), Q(\frac{3}{2}, \frac{15}{4}), P(\frac{3}{13}, \frac{24}{13}) \text{ and } S(\frac{18}{7}, \frac{2}{7}) \text{. } $

Now, evaluating the value of $Z$ for the feasible region RQPS.

Hence, the maximum value of $Z$ is 9 at $(\frac{3}{2}, \frac{15}{4})$ and the minimum value of $Z$ is $\frac{22}{7}$ at $(\frac{18}{7}, \frac{2}{7})$.

Solution

Let $x$ units of type A and $y$ units of type B electric circuits be produced by the manufacturer.

As per the given information, we construct the following table:

Now, we have the total profit in rupees $Z=50 x+60 y$ to maximise subject to the constraints $20 x+10 y \leq 200$ …(i);

$\quad 10 x+20 y \leq 120$…(ii)

$10 x+30 y \leq 150$ …(iii);

$\quad x \geq 0, y \geq 0$…(iv)

Hence, the required LPP is

Maximise $Z=50 x+60 y$ subject to the constraints

$20 x+10 y \leq 200 \Rightarrow 2 x+y \leq 20 ; 10 x+20 y \leq 120 \Rightarrow x+2 y \leq 12$

and $\quad 10 x+30 y \leq 150 \Rightarrow x+3 y \leq 15, x \geq 0, y \geq 0$

Solution

Let $x$ and $y$ be the number of large and small vans respectively. From the given information, we construct the following corresponding constraints table;

Now the objective function for minimum cost is

$ Z=400 x+200 y $

Subject to the constraints;

$$ \begin{equation*} 200 x+80 y \geq 1200 \Rightarrow 5 x+2 y \geq 30 \tag{i} \end{equation*} $$

$$ \begin{align*} 400 x+200 y & \leq 3000 \Rightarrow 2 x+y \leq 15 \tag{ii}\\ x & \leq y \tag{iii} \end{align*} $$

and $\quad x \geq 0, y \geq 0$ (non-negative constraints)

Hence, the required LPP is to minimise $Z=400 x+200 y$

Subject to the constraints $5 x+2 y \geq 30,2 x+y \leq 15, x \leq y$ and $x \geq 0, y \geq 0$.

Solution

Let the company manufactures $x$ boxes of type A screws and $y$ boxes of type B screws.

From the given information, we can construct the following table.

As per the information in the above table, the objective function for maximum profit $Z=100 x+170 y$

Subject to the constraints

$$ \begin{align*} 2 x+8 y & \leq 3600 \Rightarrow x+4 y \leq 1800 \tag{i}\\ 3 x+2 y & \leq 3600 \tag{ii}\\ x & \geq 0, y \geq 0 \end{align*} $$

Hence, the required LPP is

Maximise $\quad Z=100 x+170 y$

Subject to the constraints,

$x+4 y \leq 1800,3 x+2 y \leq 3600, x \geq 0, y \geq 0$.

Solution

Let $x$ and $y$ be the number of sweaters of type A and type B respectively.

From the given information, we have the following constraints.

$360 x+120 y \leq 72000 \Rightarrow 3 x+y \leq 600$

$x+y \leq 300 \quad \ldots$ (ii); $x+100 \geq y \Rightarrow y \leq x+100$

Profit $\quad(Z)=200 x+120 y$

Hence, the required LPP to maximise the profit is

Maximise $Z=200 x+120 y$ subject to the constraints

$3 x+y \leq 600, x+y \leq 300, y \leq x+100, x \geq 0, y \geq 0$.

Solution

Let the man covers $x km$ on his motorcycle at the speed of $50 km / hr$ and covers $y km$ at the speed of $80 km / hr$.

So, cost of petrol $=2 x+3 y$

The man has to spend ₹ 120 atmost on petrol

$\therefore \quad 2 x+3 y \leq 120$

Now, the man has only $1 hr$ time

$$ \begin{gather*} \therefore \quad \frac{x}{50}+\frac{y}{80} \leq 1 \Rightarrow 8 x+5 y \leq 400 \tag{ii}\\ x \geq 0, y \geq 0 \end{gather*} $$

To have maximum distance $Z=x+y$.

Hence, the required LPP to travel maximum distance is maximise $Z=x+y$, subject to the constraints $2 x+3 y \leq 120,8 x+5 y \leq 400, x \geq 0, y \geq 0$.

Solution

As per the solution of Question No. 11, we have Maximise $Z=50 x+60 y$ subject to the contraints $2 x+y \leq 20 \ldots(i) ; x+2 y \leq 12 \ldots$… (ii); $x+3 y \leq 15 \ldots(i i i) ; x \geq 0, y \geq 0 \ldots$… (iv)

Let us draw the table for the above statements

Table for $(i)$

$\begin{array}{|l|l|l|} \hline x & 0 & 10 \\ \hline y & 20 & 0 \\ \hline \end{array}$

Table for (ii)

$\begin{array}{|l|l|l|} \hline x & 0 & 12 \\ \hline y & 6 & 0 \\ \hline \end{array}$

Table for (ii)

$\begin{array}{|l|l|l|} \hline x & 0 & 15 \\ \hline y & 5 & 0 \\ \hline \end{array}$

Solving eq. (i) and (ii) we get,

$x=\frac{28}{3}, y=\frac{4}{3} \quad \therefore B(\frac{28}{3}, \frac{4}{3})$ is the corner

Solving eq. (ii) and (iii) we get,

$x=6, y=3 \quad \therefore C(6,3)$ is the corner

Solving eq. (i) and (iii) we get,

$x=9, y=2 \quad$ (not included in the feasible region)

Here, $OABCD$ is the feasible region.

So, the corner points are $O(0,0), A(10,0), B(\frac{28}{3}, \frac{4}{3}), C(6,3)$

and $D(0,5)$.

Let us evaluate the value of $Z$

Here, the maximum profit is ₹ 546.6 which is not possible for number of items in fraction.

Hence, the maximum profit for the manufacturer is ₹ 480 at $(6,3)$. Type $A=6$ and Type B $=3$.

Solution

As per the solution of Q. 12 ., we have $Z=400 x+200 y$ Subject to the constraints

$5 x+2 y \geq 30$

$x \leq y, x \geq 0, y \geq 0$

$x-y \leq 0$

…(i); $2 x+y \leq 15$

Let $5 x+2 y=30$

$\begin{array}{|l|l|l|} \hline x & 0 & 6 \\ \hline y & 15 & 0 \\ \hline \end{array}$

Let $2 x+y=15$

$\begin{array}{|l|l|l|} \hline x & 0 & 7.5 \\ \hline y & 15 & 0 \\ \hline \end{array}$

Let $x-y=0$

$\begin{array}{|l|l|l|} \hline x & 0 & 1 \\ \hline y & 0 & 1 \\ \hline \end{array}$

Solving eq. (i) and (iii) we get; $x=\frac{30}{7}$ and $y=\frac{30}{7}$

and on solving eq. (ii) and (iii) we get, $x=5$ and $y=5$

Here, $A B C$ is the shaded feasible region whose corner points are $A(\frac{30}{7}, \frac{30}{7}), B(5,5)$ and $C(0,15)$

Evaluating the value of $Z$, we have

Hence, the required minimum cost is $₹ 2571.4$ at $(\frac{30}{7}, \frac{30}{7})$.

Solution

As per the solution of Q. 13 , we have:

Let $3 x+2 y=3600$

$\begin{array}{|l|l|l|} \hline x & 0 & 1200 \\ \hline y & 1800 & 0 \\ \hline \end{array}$

Let $x+4 y=1800$

$\begin{array}{|l|l|l|} \hline x & 0 & 1800 \\ \hline y & 450 & 0 \\ \hline \end{array}$

Maximise $Z=100 x+170 y$ subject to the constraints

$$ \begin{align*} 3 x+2 y & \leq 3600 \quad \ldots(i) ; x+4 y \leq 1800 \tag{ii}\\ x & \geq 0, y \geq 0 \end{align*} $$

On solving eq. (i) and (ii) we get $x=1080$ and $y=180$

$OABC$ is the feasible region whose corner points are $O(0,0)$, $A(1200,0), B(1080,180), C(0,450)$.

Let us evaluate the value of $Z$.

Hence, the maximum value of $Z$ is 138600 at $(1080,180)$.

Solution

Referring to the solution of Q. 14 , we have

Maximise $\quad Z=200 x+120 y$ subject to the constraints

$$ \begin{align*} x+y & \leq 300 \quad \ldots(i) ; \quad 3 x+y \leq 600 \tag{ii}\\ x-y & \geq-100 \tag{iii}\\ x & \geq 0, y \geq 0 \end{align*} $$

On solving eq. (i) and (iii) we have $x=100, y=200$

On solving eq. (i) and (ii) we get $x=150, y=150$

Let $x+y=300$

Let $3 x+y=600 \quad$ Let $x+y=-100$

$\begin{array}{|l|l|l|} \hline x & 0 & 300 \\ \hline y & 300 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 200 \\ \hline y & 600 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & -100 \\ \hline y & 100 & 0 \\ \hline \end{array}$

Here, the shaded region is the feasible region whose corner points are $O(0,0), A(200,0), B(150,150), C(100,200), D(0,100)$. Let us evaluate the value of $Z$.

Hence, the maximum value of $Z$ is 48000 at $(150,150)$ i.e., 150 sweaters of each type.

Solution

Referring to the solution of Q.15 , we have Maximise $Z=x+y$ subject to the constraints Let $2 x+3 y=120 \quad$ Let $8 x+5 y=400$

$ \begin{aligned} 2 x+3 y & \leq 120 \ldots(i) ; 8 x+5 y \leq 400 \ldots(i i) \\ x & \geq 0, y \geq 0 \end{aligned} $

On solving eq. ( $i$ ) and (ii) we get; $x=\frac{300}{7}$ and $y=\frac{80}{7}$

Here, $OABC$ is the feasible region whose corner points are $O(0,0), A(50,0), B(\frac{300}{7}, \frac{80}{7})$ and $C(0,40)$.

Let us evaluate the value of $Z$

Hence, the maximum distance that the man can travel is $54 \frac{2}{7} km$ at $(\frac{300}{7}, \frac{80}{7})$.

Solution

We are given that $Z=x+y$ subject to the constraints

$$ \begin{align*} x+4 y & \leq 8 \tag{i}\\ 2 x+3 y & \leq 12 \tag{ii}\\ 3 x+y & \leq 9 \tag{iii}\\ x & \geq 0, y \geq 0 \end{align*} $$

$\begin{array}{|l|l|l|} \hline x & 0 & 8 \\ \hline y & 2 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 6 \\ \hline y & 4 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 3 \\ \hline y & 9 & 0 \\ \hline \end{array}$

On solving eq. (i) and (iii) we get

$x=\frac{28}{11}$ and $y=\frac{15}{11}$

Here, $OABC$ is the feasible region whose corner points are $O(0,0), A(3,0), B(\frac{28}{11}, \frac{15}{11}), C(0,2)$

Let us evaluate the value of $Z$

Hence, the maximum value of $Z$ is 3.9 at $(\frac{28}{11}, \frac{15}{11})$.

Solution

Let $x$ and $y$ be the number of Models of bike produced by the manufacturer.

Given information is

Model X takes 6 man-hours to make per unit

Model Y takes 10 man-hours to make per unit

Total man-hours available $=450$

$\therefore \quad 6 x+10 y \leq 450 \Rightarrow 3 x+5 y \leq 225$

Handling and marketing cost of Model X and Y are ₹ 2,000 and ₹ 1,000 respectively

Total funds available is ₹ 80,000 per week

$$ \begin{matrix} \therefore & 2000 x+1000 y & \leq 80,000 \\ \Rightarrow & & 2 x+y & \leq 80 \tag{ii} \end{matrix} $$

and

$ x \geq 80, y \geq 0 $

Profit (Z) per unit of models $X$ and $Y$ are $₹ 1,000$ and $₹ 500$ respectively

So,

$ Z=1000 x+500 y $

The required LPP is

Maximise $Z=1000 x+500 y$ subject to the constraints

$$ \begin{align*} 3 x+5 y & \leq 225 \tag{i}\\ 2 x+y & \leq 80 \tag{ii}\\ x & \geq 0, y \geq 0 \tag{iii} \end{align*} $$

$\begin{array}{|l|l|l|} \hline x & 0 & 40 \\ \hline y & 45 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 1 & 40 \\ \hline y & 80 & 0 \\ \hline \end{array}$

On solving eq. (i) and (i) we get, $x=25, y=30$ Here, the feasible region is $OABC$, whose corner points are $O(0,0), A(40,0), B(25,30)$ and $C(0,45)$.

Let us evaluate the value of Z.

Hence, the maximum profit is $₹ 40,000$ by producing 25 bikes of Model X and 30 bikes of Model Y.

Solution

Let there be $x$ units of tablet $X$ and $y$ units of tablet $Y$ So, according to the given information, we have

$$ \begin{gather*} 6 x+2 y \geq 18 \quad \Rightarrow \quad 3 x+y \geq 9 \tag{i}\\ 3 x+3 y \geq 21 \quad \Rightarrow \quad x+y \geq 7 \\ 2 x+4 y \geq 16 \quad \Rightarrow \quad x+2 y \geq 8 \\ x \geq 0, y \geq 0 \end{gather*} $$

$\begin{array}{|l|l|l|} \hline x & 0 & 3 \\ \hline y & 9 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 7 \\ \hline y & 7 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 8 \\ \hline y & 4 & 0 \\ \hline \end{array}$

The price of each table of $X$ type is $₹ 2$ and that of $y$ is $₹ 1$. So, the required LPP is Minimise $Z=2 x+y$ subject to the constraints $3 x+y \geq 9, x+y \geq 7, x+2 y \geq 8, x \geq 0, y \geq 0$

On solving (ii) and (iii) we get $x=6$ and $y=1$

On solving (i) and (ii) we get $x=1$ and $y=6$

From the graph, we see that the feasible region $A B C D$ is unbounded whose corner points are $A(8,0), B(6,1), C(1,6)$ and $D(0,9)$.

Let us evaluate the value of $Z$

Here, we see that 8 is the minimum value of $Z$ at $(1,6)$ but the feasible region is unbounded. So, 8 may or may not be the minimum value of $Z$.

To confirm it, we will draw a graph of inequality $2 x+y<8$ and check if it has a common point.

We see from the graph that there is no common point on the line.

Hence, the minimum value of $Z$ is 8 at $(1,6)$.

Tablet $X=1$

Table $Y=6$.

Solution

Let factory I be operated for $x$ days and II for $y$ days

At factory I: 50 calculators of model A and at factory II, 40 calculators of model A are made everyday.

Company has orders of atleast 6400 calculators of model A.

$\therefore 50 x+40 y \geq 6400 \Rightarrow 5 x+4 y \geq 640$

Also, at factory I, 50 calculators of model B and at factory II, 20 calculators of model B are made everyday.

Company has the orders of atleast 4000 of calculators of model B. $\therefore 50 x+20 y \geq 4000 \quad \Rightarrow \quad 5 x+2 y \geq 4000$

Similarly for model C,

$30 x+40 y \geq 4800 \Rightarrow 3 x+4 y \geq 480$

and $x \geq 0, y \geq 0$

It costs ₹ 12,000 and ₹ 15000 to operate the factories I and II each day.

$\therefore$ Required LPP is

Minimise $Z=12000 x+15000 y$ subject to the constraints

$$ \begin{align*} 5 x+4 y & \geq 640 \tag{i}\\ 5 x+2 y & \geq 400 \tag{ii}\\ 3 x+4 y & \geq 480 \tag{iii}\\ x & \geq 0, y \geq 0 \tag{iv} \end{align*} $$

Table for (i) equation $5 x+4 y=640$

$\begin{array}{|l|l|l|} \hline x & 0 & 128 \\ \hline y & 160 & 0 \\ \hline \end{array}$

Table for (ii) equation $5 x+2 y=400$

$\begin{array}{|l|l|l|} \hline x & 0 & 80 \\ \hline y & 200 & 0 \\ \hline \end{array}$

Table for (iii) equation $3 x+4 y=480$

$\begin{array}{|l|l|l|} \hline x & 0 & 160 \\ \hline y & 120 & 0 \\ \hline \end{array}$

On solving eq. (i) and (iii), we get $x=80, y=60$

On solving eq. (i) and (ii) we get $x=32$ and $y=120$

From the graph, we see that the feasible region $ABCD$ is open unbounded whose corners are $A(160,0), B(80,60), C(32,120)$ and $D(0,200)$.

Let us find the values of $Z$.

From the above table, it is clear that the value of $Z=1860000$ may or may not be minimum for an open unbounded region.

Now, to decide this, we draw a graph of

$ 12000 x+15000 y<1860000 $

$\Rightarrow \quad 4 x+5 y<620$

and we have to check whether there is a common point in this feasible region or not.

So, from the graph, there is no common point.

$\therefore Z=12000 x+15000 y$ has minimum value 1860000 at $(80,60)$.

Factory I : 80 days

Factory II: 60 days.

Solution

Given LPP is

Maximise and minimise $Z=3 x-4 y$ subject to

$$ \begin{matrix} x-2 y \leq 0 \qquad\text{(i)}\\ -3 x+y \leq 4 \tag{ii} \end{matrix} $$

$$ \begin{align*} & x-y \leq 6 \tag{iii}\\ & \text{ and } \quad x, y \geq 0 \tag{iv} \end{align*} $$

$\begin{array}{|l|l|l|} \hline x & 0 & 2 \\ \hline y & 0 & 1 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & -4/3 \\ \hline y & 4 & 0 \\ \hline \end{array}$

$\begin{array}{|l|l|l|} \hline x & 0 & 6 \\ \hline y & -6 & 0 \\ \hline \end{array}$

From the graph, we see that $AOB$ is open unbounded region whose corners are $O(0,0), A(0,4), B(12,6)$.

Let us evaluate the value of $Z$

For this unbounded region, the value of $Z$ may or may not be -16 . So to decide it, we draw a graph of inequality $3 x-4 y<-16$ and check whether the open half plane has common points with feasible region or not. But from the graph, we see that it has common points with the feasible region, so it will have not minimum value of $Z$. Similarly for maximum value, we draw the graph of inequality $3 x-4 y>12$ in which there is no common point with the feasible region.

Hence, the maximum value of $Z$ is 12 .

Solution

$\begin{array}{|c|c|c|} \hline \text{Corner points} & \text{Value of} Z=4 x+3 y \\ \hline(0,0) & \mathrm{Z}=0 \\ \hline(0,40) & Z=0+3(40)=120 \\ \hline(20,40) & Z=4(20)+3(40)=200 \\ \hline(60,20) & Z=4(60)+3(20)=300\rightarrow \text{Maximum} \\ \hline(60,0) & Z=4(60)+3(0)=240 \\ \hline \end{array}$

Hence, the correct option is $(b)$.

• Option (a) is incorrect: The maximum value of ( Z ) calculated from the given corner points is 300, which is less than 325. Therefore, the quantity in Column A is not greater than the quantity in Column B.

• Option (c) is incorrect: The maximum value of ( Z ) is 300, which is not equal to 325. Therefore, the two quantities are not equal.

• Option (d) is incorrect: The relationship between the quantities can be determined based on the information supplied. The maximum value of ( Z ) is clearly calculated as 300, which is less than 325. Therefore, the relationship can be determined.

Solution

Hence, the correct option is $(b)$.

• Option (a) $(0,0)$: This point gives a value of $Z=0$, which is not the minimum value of $Z$. The minimum value of $Z$ is $-32$ at point $(0,8)$.

• Option (c) $(5,0)$: This point gives a value of $Z=15$, which is higher than the minimum value of $Z$. The minimum value of $Z$ is $-32$ at point $(0,8)$.

• Option (d) $(4,10)$: This point gives a value of $Z=-28$, which is not the minimum value of $Z$. The minimum value of $Z$ is $-32$ at point $(0,8)$.

Solution

According to solution of Q. 27 , the maximum value of $Z$ is 15 at $A(5,0)$.

Hence, the correct option is $(a)$.

• Option (b) $(6,5)$: This point does not yield the maximum value of $Z$ as determined in Exercise 27. The maximum value of $Z$ is 15 at $(5,0)$, not at $(6,5)$.

• Option (c) $(6,8)$: This point does not yield the maximum value of $Z$ as determined in Exercise 27. The maximum value of $Z$ is 15 at $(5,0)$, not at $(6,8)$.

• Option (d) $(4,10)$: This point does not yield the maximum value of $Z$ as determined in Exercise 27. The maximum value of $Z$ is 15 at $(5,0)$, not at $(4,10)$.

Solution

According to the solution of Q. 27 , Maximum value of $Z=15$ and Minimum value of $Z=-32$

So, the sum of Maximum value and Minimum value of $Z$

$ =15+(-32)=-17 $

Hence, the correct option is $(d)$.

• Option (a) 13 is incorrect because the sum of the maximum value of $Z$ (15) and the minimum value of $Z$ (-32) is not equal to 13. The correct sum is -17.

• Option (b) 1 is incorrect because the sum of the maximum value of $Z$ (15) and the minimum value of $Z$ (-32) is not equal to 1. The correct sum is -17.

• Option (c) -13 is incorrect because the sum of the maximum value of $Z$ (15) and the minimum value of $Z$ (-32) is not equal to -13. The correct sum is -17.

Solution

The feasible region is shown in the figure for which the objective function $F=3 x-4 y$

Hence, the correct option is (c).

• Option (a) is incorrect because the maximum value of ( F ) is not 0. The value of ( F ) at the origin ( O(0,0) ) is 0, but this is not the maximum value within the feasible region.

• Option (b) is incorrect because the maximum value of ( F ) is not 8. The value of ( F ) at the corner points ( O(0,0) ), ( A(12,6) ), and ( B(0,4) ) are 0, 12, and -16 respectively. None of these values are 8.

• Option (d) is incorrect because the maximum value of ( F ) is not -18. The value of ( F ) at the corner points ( O(0,0) ), ( A(12,6) ), and ( B(0,4) ) are 0, 12, and -16 respectively. None of these values are -18.

Solution

According to the solution of Q.30, the minimum value of $F$ is -16 at $(0,4)$.

Hence, the correct option is (b).

• Option (a) is incorrect because the minimum value of ( F ) is not 0; it is -16 at the point (0,4).
• Option (c) is incorrect because the minimum value of ( F ) is not 12; it is -16 at the point (0,4).
• Option (d) is incorrect because the minimum value of ( F ) does exist and is -16 at the point (0,4).

Solution

The minimum value of $F$ occurs at any point on the line segment joining the points $(0,2)$ and $(3,0)$.

Hence, the correct option is $(d)$.

• Option (a) $(0,2)$ only: This option is incorrect because the minimum value of $F$ is not unique to the point $(0,2)$. The value of $F$ at $(0,2)$ is 12, but the same value is also obtained at $(3,0)$. Therefore, the minimum value occurs at multiple points, not just at $(0,2)$.

• Option (b) $(3,0)$ only: This option is incorrect because the minimum value of $F$ is not unique to the point $(3,0)$. The value of $F$ at $(3,0)$ is 12, but the same value is also obtained at $(0,2)$. Therefore, the minimum value occurs at multiple points, not just at $(3,0)$.

• Option (c) the mid-point of the line segment joining the points $(0,2)$ and $(3,0)$ only: This option is incorrect because the minimum value of $F$ is not unique to the midpoint of the line segment joining $(0,2)$ and $(3,0)$. The minimum value of $F$ occurs at any point on the entire line segment joining $(0,2)$ and $(3,0)$, not just at the midpoint.

Solution

According to the solution of Q. 32 ,

Maximum value of $F-$ Minimum value of $F=72-12=60$

Hence, the correct option is $(a)$.

• Option (b) 48 is incorrect because the difference between the maximum and minimum values of $F$ is 60, not 48.
• Option (c) 42 is incorrect because the difference between the maximum and minimum values of $F$ is 60, not 42.
• Option (d) 18 is incorrect because the difference between the maximum and minimum values of $F$ is 60, not 18.

Solution

So, condition of $p$ and $q$ so that the minimum of $Z$ occurs at $(3,0)$ and $(1,1)$ is

$p+q=3 p \Rightarrow p-3 p+q=0 \Rightarrow p=\frac{q}{2}$.

Hence, the correct option is (b).

• Option (a) $p=2q$: If $p=2q$, then the value of $Z$ at the corner points would be:

  • At $(0,3)$: $Z = 3q$
  • At $(1,1)$: $Z = 2q + q = 3q$
  • At $(3,0)$: $Z = 3 \cdot 2q = 6q$ The minimum value of $Z$ would be $3q$ at both $(0,3)$ and $(1,1)$, not at $(3,0)$ and $(1,1)$.

• Option (c) $p=3q$: If $p=3q$, then the value of $Z$ at the corner points would be:

  • At $(0,3)$: $Z = 3q$
  • At $(1,1)$: $Z = 3q + q = 4q$
  • At $(3,0)$: $Z = 3 \cdot 3q = 9q$ The minimum value of $Z$ would be $3q$ at $(0,3)$, not at $(3,0)$ and $(1,1)$.

• Option (d) $p=q$: If $p=q$, then the value of $Z$ at the corner points would be:

  • At $(0,3)$: $Z = 3q$
  • At $(1,1)$: $Z = q + q = 2q$
  • At $(3,0)$: $Z = 3q$ The minimum value of $Z$ would be $2q$ at $(1,1)$, not at $(3,0)$ and $(1,1)$.

Solution

constraints.

Solution

linear.

Solution

open unbounded

Solution

maximum

Solution

bounded

Solution

intersection

Solution

convex

Solution

True.

Solution

False.

Solution

False.

Solution

True.