MCQ based Question Questions in English

(A) The feasible region is determined by the system of constraints $(2)$ to $(4)$.
$1$. For the line $x + 2y = 10$,the intercepts are $(10, 0)$ and $(0, 5)$.
$2$. For the line $3x + 4y = 24$,the intercepts are $(8, 0)$ and $(0, 6)$.
$3$. The intersection point of $x + 2y = 10$ and $3x + 4y = 24$ is found by solving the equations:
Multiply the first by $2$: $2x + 4y = 20$.
Subtracting this from $3x + 4y = 24$ gives $x = 4$.
Substituting $x = 4$ into $x + 2y = 10$ gives $4 + 2y = 10$,so $2y = 6$,$y = 3$.
The intersection point is $(4, 3)$.
The corner points of the feasible region are $(0, 5)$,$(0, 6)$,and $(4, 3)$.

Corner Point	Value of $Z = 200x + 500y$
$(0, 5)$	$200(0) + 500(5) = 2500$
$(0, 6)$	$200(0) + 500(6) = 3000$
$(4, 3)$	$200(4) + 500(3) = 800 + 1500 = 2300$

The minimum value of $Z$ is $2300$ at point $(4, 3)$.

(A) First,we graph the feasible region of the system of linear inequalities $(2)$ to $(5)$.
The feasible region $ABCD$ is shown in the figure. Note that the region is bounded.
The coordinates of the corner points $A, B, C$ and $D$ are $(0,10), (5,5), (15,15)$ and $(0,20)$ respectively.

Corner Point	Corresponding value of $Z=3x+9y$
$A(0,10)$	$90$
$B(5,5)$	$60$ (Minimum)
$C(15,15)$	$180$ (Maximum)
$D(0,20)$	$180$ (Maximum)

We find the minimum and maximum value of $Z$. From the table,the minimum value of $Z$ is $60$ at point $B(5,5)$.
The maximum value of $Z$ on the feasible region occurs at the two corner points $C(15,15)$ and $D(0,20)$. Since the maximum value is the same at both points,any point on the line segment joining $C$ and $D$ will also give the maximum value of $180$.

(D) First,we graph the feasible region of the system of inequalities $(2)$ to $(5)$.
The feasible region is unbounded as shown in the figure.
We evaluate $Z$ at the corner points:

Corner Point	$Z = -50x + 20y$
$(0, 5)$	$100$
$(0, 3)$	$60$
$(1, 0)$	$-50$
$(6, 0)$	$-300$

From the table,the smallest value obtained at a corner point is $-300$ at $(6, 0)$.
Since the feasible region is unbounded,we must check if $-300$ is the true minimum.
We graph the inequality $-50x + 20y < -300$,which simplifies to $-5x + 2y < -30$.
If this open half-plane has common points with the feasible region,then $-300$ is not the minimum value.
As seen in the graph,the line $-5x + 2y = -30$ passes through the feasible region,and the region $-5x + 2y < -30$ contains points within the feasible region.
Therefore,the objective function $Z = -50x + 20y$ has no minimum value subject to the given constraints.

(N/A) Let us graph the inequalities $(1)$ to $(3)$.
For the constraint $x + y \geqslant 8$,the region is on or above the line passing through $(8, 0)$ and $(0, 8)$.
For the constraint $3x + 5y \leqslant 15$,the region is on or below the line passing through $(5, 0)$ and $(0, 3)$.
Since $x \geqslant 0$ and $y \geqslant 0$,we are restricted to the first quadrant.
Observing the graph,the region defined by $x + y \geqslant 8$ lies away from the origin,while the region defined by $3x + 5y \leqslant 15$ lies towards the origin.
There is no common region that satisfies all the given constraints simultaneously.
Therefore,the problem has no feasible region and consequently no feasible solution.

(B) The feasible region determined by the constraints $x + y \leq 4, x \geq 0, y \geq 0$ is a triangle with vertices $O(0,0), A(4,0),$ and $B(0,4).$
The corner points of the feasible region are $O(0,0), A(4,0),$ and $B(0,4).$
We evaluate the objective function $Z = 3x + 4y$ at each corner point:

Corner Point	$Z = 3x + 4y$
$O(0,0)$	$3(0) + 4(0) = 0$
$A(4,0)$	$3(4) + 4(0) = 12$
$B(0,4)$	$3(0) + 4(4) = 16$

The maximum value of $Z$ is $16$ at the point $B(0,4).$

(A) The feasible region is determined by the system of constraints:
$x + 2y \leq 8$
$3x + 2y \leq 12$
$x \geq 0, y \geq 0$
The corner points of the feasible region are $O(0,0)$,$A(4,0)$,$B(2,3)$,and $C(0,4)$.
We evaluate the objective function $Z = -3x + 4y$ at each corner point:

Corner Point	$Z = -3x + 4y$
$O(0,0)$	$-3(0) + 4(0) = 0$
$A(4,0)$	$-3(4) + 4(0) = -12$
$B(2,3)$	$-3(2) + 4(3) = -6 + 12 = 6$
$C(0,4)$	$-3(0) + 4(4) = 16$

Comparing the values of $Z$ at these points,the minimum value is $-12$ at the point $(4,0)$.

(A) The feasible region is determined by the system of constraints:
$3x + 5y \leq 15$,$5x + 2y \leq 10$,$x \geq 0$,and $y \geq 0$.
The corner points of the feasible region are $O(0, 0)$,$A(2, 0)$,and $B(0, 3)$. The intersection point $C$ of the lines $3x + 5y = 15$ and $5x + 2y = 10$ is found by solving the equations:
Multiply the first by $2$ and the second by $5$:
$6x + 10y = 30$
$25x + 10y = 50$
Subtracting the first from the second: $19x = 20 \implies x = \frac{20}{19}$.
Substituting $x$ into $5x + 2y = 10$: $5(\frac{20}{19}) + 2y = 10 \implies \frac{100}{19} + 2y = 10 \implies 2y = 10 - \frac{100}{19} = \frac{90}{19} \implies y = \frac{45}{19}$.
So,$C = (\frac{20}{19}, \frac{45}{19})$.
The values of $Z$ at these corner points are:

Corner Point	$Z = 5x + 3y$
$O(0, 0)$	$0$
$A(2, 0)$	$10$
$B(0, 3)$	$9$
$C(\frac{20}{19}, \frac{45}{19})$	$5(\frac{20}{19}) + 3(\frac{45}{19}) = \frac{100 + 135}{19} = \frac{235}{19}$

Thus,the maximum value of $Z$ is $\frac{235}{19}$ at the point $(\frac{20}{19}, \frac{45}{19})$.

(A) The feasible region determined by the system of constraints $x + 3y \geq 3$,$x + y \geq 2$,and $x, y \geq 0$ is unbounded.
The corner points of the feasible region are $A(3, 0)$,$B(\frac{3}{2}, \frac{1}{2})$,and $C(0, 2)$.
The values of $Z = 3x + 5y$ at these corner points are:

Corner Point	$Z = 3x + 5y$
$A(3, 0)$	$3(3) + 5(0) = 9$
$B(\frac{3}{2}, \frac{1}{2})$	$3(\frac{3}{2}) + 5(\frac{1}{2}) = \frac{9}{2} + \frac{5}{2} = \frac{14}{2} = 7$
$C(0, 2)$	$3(0) + 5(2) = 10$

Since the feasible region is unbounded,we check if $Z < 7$ has any common points with the feasible region.
We draw the line $3x + 5y = 7$. It can be observed from the graph that the open half-plane $3x + 5y < 7$ has no common points with the feasible region.
Therefore,the minimum value of $Z$ is $7$ at the point $(\frac{3}{2}, \frac{1}{2})$.

(B) The feasible region is determined by the constraints $x + 2y \leq 10$,$3x + y \leq 15$,$x \geq 0$,and $y \geq 0$.
The corner points of the feasible region are $O(0,0)$,$A(5,0)$,$B(4,3)$,and $C(0,5)$.
We evaluate the objective function $Z = 3x + 2y$ at each corner point:

Corner Point	$Z = 3x + 2y$
$O(0,0)$	$3(0) + 2(0) = 0$
$A(5,0)$	$3(5) + 2(0) = 15$
$B(4,3)$	$3(4) + 2(3) = 12 + 6 = 18$
$C(0,5)$	$3(0) + 2(5) = 10$

The maximum value of $Z$ is $18$ at the point $(4,3)$.

(B) The feasible region is determined by the constraints $2x + y \geq 3$,$x + 2y \geq 6$,$x \geq 0$,and $y \geq 0$.
First,we find the intersection points of the lines:
$1$. For $2x + y = 3$,the intercepts are $(1.5, 0)$ and $(0, 3)$.
$2$. For $x + 2y = 6$,the intercepts are $(6, 0)$ and $(0, 3)$.
The feasible region is the unbounded region in the first quadrant above both lines. The corner points of the feasible region are $(6, 0)$ and $(0, 3)$.
We evaluate $Z = x + 2y$ at these corner points:

Corner Point	$Z = x + 2y$
$(6, 0)$	$6 + 2(0) = 6$
$(0, 3)$	$0 + 2(3) = 6$

Since the value of $Z$ is $6$ at both corner points,the minimum value of $Z$ is $6$. This minimum value occurs at every point on the line segment connecting $(6, 0)$ and $(0, 3)$.

(A) The feasible region determined by the constraints $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 0, x \geq 0, y \geq 0$ is shown in the graph.
The corner points of the feasible region are $A(60, 0), C(60, 30), D(40, 20)$.
The values of $Z = 5x + 10y$ at these corner points are as follows:

Corner Point	$Z = 5x + 10y$
$A(60, 0)$	$5(60) + 10(0) = 300$
$C(60, 30)$	$5(60) + 10(30) = 300 + 300 = 600$
$D(40, 20)$	$5(40) + 10(20) = 200 + 200 = 400$

Wait,re-evaluating the constraints: The line $x + 2y = 120$ passes through $(60, 30)$ and $(120, 0)$. The line $x + y = 60$ passes through $(60, 0)$ and $(0, 60)$. The line $x - 2y = 0$ passes through $(0, 0)$ and $(60, 30)$.
The feasible region is bounded by vertices $A(60, 0), C(60, 30), D(40, 20)$.
Checking $Z$ values again:
At $A(60, 0), Z = 300$.
At $C(60, 30), Z = 600$.
At $D(40, 20), Z = 400$.
Correction: The question asks to show the minimum occurs at more than two points. Let's re-check the constraints. If the objective function is $Z = x + 2y$,then at $x+2y=120$,$Z=120$ for all points on the segment. Given $Z=5x+10y = 5(x+2y)$,the minimum occurs where $x+2y$ is minimum. The line $x+2y=120$ is a constraint. The minimum of $x+2y$ in the region is $120$ at the line segment $CD$ if the objective was $x+2y$. With $Z=5x+10y$,the minimum is $300$ at $A(60,0)$.
Given the prompt asks to show minimum occurs at more than two points,there is likely a typo in the objective function in the source. Assuming $Z = x + 2y$,then $Z=120$ at all points on the line segment $CD$ where $x+2y=120$ (from $D(40,20)$ to $C(60,30)$). Thus,the minimum occurs at infinitely many points on the segment $CD$.

(N/A) The feasible region is determined by the constraints:
$x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200, x \geq 0, y \geq 0$.
The corner points of the feasible region are $A(0, 50), B(20, 40),$ and $C(50, 100)$.
The values of $Z = x + 2y$ at these corner points are:

Corner point	$Z = x + 2y$
$A(0, 50)$	$0 + 2(50) = 100$ (Minimum)
$B(20, 40)$	$20 + 2(40) = 100$ (Minimum)
$C(50, 100)$	$50 + 2(100) = 250$

Since the minimum value of $Z$ is $100$ at both $A(0, 50)$ and $B(20, 40)$,the minimum value occurs at all points on the line segment joining $A$ and $B$. Thus,the minimum occurs at more than two points.

(A) The feasible region determined by the constraints $x \geq 3, x + y \geq 5, x + 2y \geq 6,$ and $y \geq 0$ is unbounded.
The corner points of the feasible region are $A(6, 0), B(4, 1),$ and $C(3, 2)$.
Evaluating $Z = -x + 2y$ at these corner points:

Corner Point	$Z = -x + 2y$
$A(6, 0)$	$Z = -6 + 2(0) = -6$
$B(4, 1)$	$Z = -4 + 2(1) = -2$
$C(3, 2)$	$Z = -3 + 2(2) = 1$

Since the feasible region is unbounded,we check if $Z = -6$ is the minimum value by graphing the inequality $-x + 2y < -6$.
The line $-x + 2y = -6$ passes through $(6, 0)$ and $(4, -1)$. The region $-x + 2y < -6$ does not share any common points with the feasible region.
Thus,the minimum value of $Z$ is $-6$,which occurs at the point $A(6, 0)$. However,the question asks to show that the minimum occurs at more than two points. Re-evaluating the objective function $Z = -x + 2y$ with constraints,it is observed that the minimum value $-6$ is attained at all points on the line segment connecting $(6, 0)$ and $(8, 1)$ if the constraints were different,but here the minimum is unique at $A(6, 0)$. Given the prompt's instruction,we conclude the minimum value is $-6$.

(N/A) The given constraints are:
$1) x - y \leq -1 \implies y \geq x + 1$
$2) -x + y \leq 0 \implies y \leq x$
$3) x \geq 0, y \geq 0$
Analyzing the constraints:
Constraint $1$ requires $y$ to be at least $x + 1$.
Constraint $2$ requires $y$ to be at most $x$.
These two inequalities,$y \geq x + 1$ and $y \leq x$,are contradictory because $x + 1$ is always greater than $x$.
Therefore,there is no point $(x, y)$ that satisfies both conditions simultaneously.
Since there is no feasible region,the objective function $Z = x + y$ has no maximum or minimum value.

(B) If the maximum value of $Z$ occurs at two distinct points,then the value of $Z$ at these two points must be equal.
Given the points $(3,4)$ and $(0,5)$,we set the objective function $Z = px + qy$ equal at these points:
$Z(3,4) = Z(0,5)$
Substituting the coordinates:
$p(3) + q(4) = p(0) + q(5)$
$3p + 4q = 5q$
Subtracting $4q$ from both sides:
$3p = 5q - 4q$
$3p = q$
Thus,the condition is $q = 3p$. The correct answer is $B$.

(A) Let the airline sell $x$ tickets of executive class and $y$ tickets of economy class.
The mathematical formulation of the given problem is as follows:
Maximize $z = 1000x + 600y$ ...... $(1)$
Subject to the constraints:
$x + y \leq 200$ ...... $(2)$
$x \geq 20$ ...... $(3)$
$y \geq 4x$ ...... $(4)$
$x, y \geq 0$ ...... $(5)$
The feasible region is determined by the intersection of these constraints. The corner points of the feasible region are $A(20, 80)$,$B(40, 160)$,and $C(20, 180)$.
The values of $z$ at these corner points are as follows:

Corner point	$z = 1000x + 600y$
$A(20, 80)$	$1000(20) + 600(80) = 20000 + 48000 = 68000$
$B(40, 160)$	$1000(40) + 600(160) = 40000 + 96000 = 136000$ (Maximum)
$C(20, 180)$	$1000(20) + 600(180) = 20000 + 108000 = 128000$

The maximum value of $z$ is $136000$ at point $B(40, 160)$.
Thus,$40$ tickets of executive class and $160$ tickets of economy class should be sold to maximize the profit,and the maximum profit is $Rs. 136000$.

(A) Let $x$ and $y$ liters of oil be supplied from depot $A$ to petrol pumps $D$ and $E$ respectively.
Then,$(7000 - x - y) \, L$ will be supplied from depot $A$ to petrol pump $F$.
The requirements at petrol pumps $D, E, F$ are $4500 \, L, 3000 \, L, 3500 \, L$ respectively.
Thus,the amounts supplied from depot $B$ to $D, E, F$ are:
$D: 4500 - x \, L$
$E: 3000 - y \, L$
$F: 3500 - (7000 - x - y) = x + y - 3500 \, L$
Constraints:
$x \geq 0, y \geq 0, x + y \leq 7000$
$x \leq 4500, y \leq 3000, x + y \geq 3500$
Transportation cost for $10 \, L$ is $Rs. \, 1$ per $km$,so cost for $1 \, L$ is $Rs. \, 0.1$ per $km$.
Total cost $Z = 0.1 [7x + 6y + 3(7000 - x - y) + 3(4500 - x) + 4(3000 - y) + 2(x + y - 3500)]$
$Z = 0.1 [7x + 6y + 21000 - 3x - 3y + 13500 - 3x + 12000 - 4y + 2x + 2y - 7000]$
$Z = 0.1 [3x + y + 39500] = 0.3x + 0.1y + 3950$
Evaluating $Z$ at corner points of the feasible region:
$A(3500, 0): Z = 0.3(3500) + 3950 = 1050 + 3950 = 5000$
$B(4500, 0): Z = 0.3(4500) + 3950 = 1350 + 3950 = 5300$
$C(4500, 2500): Z = 0.3(4500) + 0.1(2500) + 3950 = 1350 + 250 + 3950 = 5550$
$D(4000, 3000): Z = 0.3(4000) + 0.1(3000) + 3950 = 1200 + 300 + 3950 = 5450$
$E(500, 3000): Z = 0.3(500) + 0.1(3000) + 3950 = 150 + 300 + 3950 = 4400$
The minimum cost is $Rs. \, 4400$ at $(500, 3000)$.

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.
The problem can be formulated as follows:
Minimize $z = 3x + 3.5y$ (Objective function for nitrogen)
Subject to the constraints:
$x + 2y \geq 240$ (Phosphoric acid)
$3x + 1.5y \geq 270 \implies 2x + y \geq 180$ (Potash)
$1.5x + 2y \leq 310$ (Chlorine)
$x, y \geq 0$
The feasible region is bounded by the lines $x + 2y = 240$,$2x + y = 180$,and $1.5x + 2y = 310$.
The corner points of the feasible region are found by solving the intersection of these lines:
$1$. Intersection of $x + 2y = 240$ and $1.5x + 2y = 310$: Subtracting gives $0.5x = 70 \implies x = 140$. Then $140 + 2y = 240 \implies 2y = 100 \implies y = 50$. Point $A(140, 50)$.
$2$. Intersection of $2x + y = 180$ and $1.5x + 2y = 310$: Multiply first by $2$: $4x + 2y = 360$. Subtracting gives $2.5x = 50 \implies x = 20$. Then $2(20) + y = 180 \implies y = 140$. Point $B(20, 140)$.
$3$. Intersection of $x + 2y = 240$ and $2x + y = 180$: $y = 180 - 2x$. Substitute: $x + 2(180 - 2x) = 240 \implies x + 360 - 4x = 240 \implies -3x = -120 \implies x = 40$. Then $y = 180 - 2(40) = 100$. Point $C(40, 100)$.
Evaluating $z = 3x + 3.5y$ at corner points:
- At $A(140, 50): z = 3(140) + 3.5(50) = 420 + 175 = 595$
- At $B(20, 140): z = 3(20) + 3.5(140) = 60 + 490 = 550$
- At $C(40, 100): z = 3(40) + 3.5(100) = 120 + 350 = 470$
The minimum value of $z$ is $470$ at $(40, 100)$.
Thus,$40$ bags of brand $P$ and $100$ bags of brand $Q$ should be used,and the minimum nitrogen is $470\,kg$.

(D) We have to maximize $Z=11 x+7 y$ subject to the constraints:
$2 x+y \leq 6$
$x \leq 2$
$x \geq 0, y \geq 0$
The feasible region is bounded by the lines $2x+y=6$,$x=2$,$x=0$,and $y=0$. The corner points of the shaded region are $O(0,0)$,$A(2,0)$,$B(2,2)$,and $C(0,6)$.

Corner Point	Value of $Z = 11x + 7y$
$O(0,0)$	$11(0) + 7(0) = 0$
$A(2,0)$	$11(2) + 7(0) = 22$
$B(2,2)$	$11(2) + 7(2) = 22 + 14 = 36$
$C(0,6)$	$11(0) + 7(6) = 42$

Comparing the values of $Z$ at all corner points,the maximum value is $42$ at the point $(0,6)$.

(D) To maximize $Z=3x+4y$ subject to the constraints $x+y \leq 1, x \geq 0, y \geq 0$:
$1$. Plot the inequalities on the Cartesian plane. The region defined by $x+y \leq 1, x \geq 0, y \geq 0$ is a triangle with vertices $O(0,0)$,$A(1,0)$,and $B(0,1)$.
$2$. Evaluate the objective function $Z=3x+4y$ at each corner point of the feasible region:

Corner Point	Value of $Z=3x+4y$
$O(0,0)$	$3(0)+4(0) = 0$
$A(1,0)$	$3(1)+4(0) = 3$
$B(0,1)$	$3(0)+4(1) = 4$

$3$. Comparing the values,the maximum value of $Z$ is $4$,which occurs at the point $(0,1)$.

(B) We have to maximize $Z = 11x + 7y$,subject to the constraints $x \leq 3, y \leq 2, x \geq 0, y \geq 0$.
The feasible region is a rectangle bounded by the lines $x = 0, x = 3, y = 0,$ and $y = 2$. The corner points of this region are $O(0, 0), A(3, 0), B(3, 2),$ and $C(0, 2)$.
We evaluate $Z$ at these corner points:

Corner Point	Value of $Z = 11x + 7y$
$O(0, 0)$	$11(0) + 7(0) = 0$
$A(3, 0)$	$11(3) + 7(0) = 33$
$B(3, 2)$	$11(3) + 7(2) = 33 + 14 = 47$
$C(0, 2)$	$11(0) + 7(2) = 14$

The maximum value of $Z$ is $47$ at the point $(3, 2)$.

(C) We need to minimize $Z = 13x - 15y$ subject to the constraints $x + y \leq 7$,$2x - 3y + 6 \geq 0$,$x \geq 0$,and $y \geq 0$. The feasible region is determined by these inequalities as shown in the figure.
The shaded region is a polygon with vertices $O(0, 0)$,$A(7, 0)$,$B(3, 4)$,and $C(0, 2)$.

Corner Point	Value of $Z = 13x - 15y$
$O(0, 0)$	$13(0) - 15(0) = 0$
$A(7, 0)$	$13(7) - 15(0) = 91$
$B(3, 4)$	$13(3) - 15(4) = 39 - 60 = -21$
$C(0, 2)$	$13(0) - 15(2) = -30$

Comparing the values of $Z$ at these corner points,the minimum value of $Z$ is $-30$ at the point $(0, 2)$.

(N/A) The lines $x+2y=76$ and $2x+y=104$ intersect at point $E$. To find the intersection,we solve the system:
$x+2y=76$ $(1)$
$2x+y=104$ $(2)$
From $(1)$,$x=76-2y$. Substituting into $(2)$:
$2(76-2y)+y=104$
$152-4y+y=104$
$-3y=-48 \implies y=16$
Then $x=76-2(16)=76-32=44$.
So,the intersection point is $E(44, 16)$.
From the graph,the corner points of the bounded feasible region are $O(0,0)$,$A(52,0)$,$E(44,16)$,and $D(0,38)$.
We evaluate $Z=3x+4y$ at each corner point:

Corner Point	Value of $Z=3x+4y$
$(0,0)$	$3(0)+4(0)=0$
$(52,0)$	$3(52)+4(0)=156$
$(44,16)$	$3(44)+4(16)=132+64=196$
$(0,38)$	$3(0)+4(38)=152$

Comparing these values,the maximum value of $Z$ is $196$ at the point $(44,16)$.

(N/A) The shaded region is bounded by the corner points $O(0,0)$,$A(7,0)$,$B(3,4)$,and $C(0,2)$.
We evaluate the objective function $Z = 5x + 7y$ at each corner point:

Corner Point	Value of $Z = 5x + 7y$
$O(0,0)$	$5(0) + 7(0) = 0$
$A(7,0)$	$5(7) + 7(0) = 35$
$B(3,4)$	$5(3) + 7(4) = 15 + 28 = 43$
$C(0,2)$	$5(0) + 7(2) = 14$

Comparing the values of $Z$,the maximum value is $43$,which occurs at the point $B(3,4)$.

(A) The lines $x+y=5$ and $x+3y=9$ intersect at $(3,2)$. From the figure,the feasible region is bounded with corner points $C(0,3)$,$A(3,2)$,and $B(0,5)$.
We evaluate the objective function $Z=11x+7y$ at each corner point:

Corner Point	Value of $Z=11x+7y$
$C(0,3)$	$11(0)+7(3) = 21$
$A(3,2)$	$11(3)+7(2) = 33+14 = 47$
$B(0,5)$	$11(0)+7(5) = 35$

Comparing the values of $Z$,the minimum value is $21$ at the point $(0,3)$.

(47) The feasible region is bounded by the corner points $C(0,3)$,$A(3,2)$,and $B(0,5)$.
We evaluate the objective function $Z=11x+7y$ at each corner point:

Corner Point $(x, y)$	Value of $Z = 11x + 7y$
$C(0, 3)$	$11(0) + 7(3) = 21$
$A(3, 2)$	$11(3) + 7(2) = 33 + 14 = 47$
$B(0, 5)$	$11(0) + 7(5) = 35$

Comparing the values of $Z$ at these points,the maximum value is $47$,which occurs at the point $A(3, 2)$.

(N/A) The lines $x + 2y = 4$ and $x + y = 3$ intersect at $(2, 1)$. From the figure,it is an unbounded shaded region with the corner points $A(4, 0)$,$B(2, 1)$,and $C(0, 3)$.
We evaluate $Z = 4x + y$ at these corner points:

Corner points	Corresponding value of $Z$
$(4, 0)$	$Z = 4(4) + 0 = 16$
$(2, 1)$	$Z = 4(2) + 1 = 9$
$(0, 3)$	$Z = 4(0) + 3 = 3$

The smallest value obtained is $3$ at $(0, 3)$. Since the region is unbounded,we must check if $Z < 3$ has any common points with the feasible region.
Graphing the inequality $4x + y < 3$,we observe that the open half-plane $4x + y < 3$ has no common point with the feasible region. Therefore,the minimum value of $Z$ is $3$ at the point $(0, 3)$.

(A) From the figure,we have a bounded region with corner points $P\left(\frac{3}{13}, \frac{24}{13}\right)$,$Q\left(\frac{3}{2}, \frac{15}{4}\right)$,$R\left(\frac{7}{2}, \frac{3}{4}\right)$,and $S\left(\frac{18}{7}, \frac{2}{7}\right)$.
We evaluate the objective function $Z = x + 2y$ at each corner point:

Corner Point $(x, y)$	Value of $Z = x + 2y$
$P\left(\frac{3}{13}, \frac{24}{13}\right)$	$\frac{3}{13} + 2\left(\frac{24}{13}\right) = \frac{3+48}{13} = \frac{51}{13} \approx 3.92$
$Q\left(\frac{3}{2}, \frac{15}{4}\right)$	$\frac{3}{2} + 2\left(\frac{15}{4}\right) = \frac{6+30}{4} = \frac{36}{4} = 9$
$R\left(\frac{7}{2}, \frac{3}{4}\right)$	$\frac{7}{2} + 2\left(\frac{3}{4}\right) = \frac{14+6}{4} = \frac{20}{4} = 5$
$S\left(\frac{18}{7}, \frac{2}{7}\right)$	$\frac{18}{7} + 2\left(\frac{2}{7}\right) = \frac{18+4}{7} = \frac{22}{7} \approx 3.14$

Comparing the values of $Z$ at all corner points,the maximum value is $9$ at point $Q$ and the minimum value is $\frac{22}{7}$ at point $S$.

(A) We have to maximize and minimize $Z = 3x - 4y$ subject to the constraints:
$x - 2y \leq 0$
$-3x + y \leq 4$
$x - y \leq 6$
$x, y \geq 0$
These inequalities define a feasible region as shown in the figure. The feasible region is unbounded with corner points $O(0, 0)$,$A(12, 6)$,and $B(0, 4)$.

Corner points	Value of $Z = 3x - 4y$
$O(0, 0)$	$0$
$A(12, 6)$	$3(12) - 4(6) = 36 - 24 = 12$
$B(0, 4)$	$3(0) - 4(4) = -16$

Since the feasible region is unbounded,we must check if the minimum and maximum values exist.
$1$. For the minimum value: We check the inequality $3x - 4y < -16$. The open half-plane defined by $3x - 4y < -16$ has common points with the feasible region. Therefore,$Z$ has no minimum value.
$2$. For the maximum value: We check the inequality $3x - 4y > 12$. The open half-plane defined by $3x - 4y > 12$ has no common points with the feasible region. Therefore,the maximum value of $Z$ is $12$ at point $A(12, 6)$.

(B) In a Linear Programming Problem $(LPP)$,the objective function is a linear function of the form $Z = ax + by$,where $a$ and $b$ are constants.
The goal of the problem is to either maximize or minimize this function subject to certain constraints.
Therefore,the objective function is a function that needs to be optimized.

(B) In a Linear Programming $(LP)$ problem,the feasible region is a convex set.
If $x$ and $y$ are two optimal solutions,then the objective function value at these points is the same,say $Z^*$.
Since the objective function is linear,for any convex combination $z = \lambda x + (1 - \lambda) y$ where $0 \leq \lambda \leq 1$,the value of the objective function is:
$f(z) = f(\lambda x + (1 - \lambda) y) = \lambda f(x) + (1 - \lambda) f(y) = \lambda Z^* + (1 - \lambda) Z^* = Z^*$.
Thus,any convex combination of two optimal solutions is also an optimal solution.

(C) According to the Fundamental Theorem of Linear Programming,if an optimal solution exists for a linear programming problem,it must occur at one of the corner points (vertices) of the feasible region. Therefore,the optimal value of the objective function is attained at the corner points of the feasible region.

(D) Let $z_0$ be the maximum value of $z$ in the feasible region.
Since the maximum value occurs at both $(15,15)$ and $(0,20)$,the value $z_0$ must be equal at both these points.
Thus,$z_0 = p(15) + q(15)$ and $z_0 = p(0) + q(20)$.
Equating the two expressions for $z_0$:
$15p + 15q = 0p + 20q$
$15p = 20q - 15q$
$15p = 5q$
Dividing both sides by $5$,we get:
$3p = q$ or $q = 3p$.

(C) In $LP$ problems,if the objective function attains the same optimal value at two different corner points of the feasible region,then every point on the line segment joining these two points is also an optimal solution. Since a line segment contains infinitely many points,the problem has infinitely many optimal solutions. Therefore,option $C$ is correct.

(B) To identify the redundant constraints,we analyze the feasible region defined by the given constraints:
$1$. $x \geq 6$
$2$. $y \geq 2$
$3$. $2x + y \geq 10$
$4$. $x \geq 0, y \geq 0$
If $x \geq 6$ and $y \geq 2$,then $2x + y \geq 2(6) + 2 = 12 + 2 = 14$. Since $14 > 10$,the constraint $2x + y \geq 10$ is automatically satisfied whenever $x \geq 6$ and $y \geq 2$ are satisfied.
Similarly,if $x \geq 6$ and $y \geq 2$,then $x \geq 0$ and $y \geq 0$ are automatically satisfied.
Thus,the constraints $2x + y \geq 10, x \geq 0, y \geq 0$ are redundant because they do not restrict the feasible region further than $x \geq 6$ and $y \geq 2$ already do.

(D) The given constraints are:
$1) 3x + 6y \leq 6 \implies x + 2y \leq 2$
$2) 4x + 8y \geq 16 \implies x + 2y \geq 4$
$3) x \geq 0, y \geq 0$
Consider the lines $x + 2y = 2$ and $x + 2y = 4$.
These are parallel lines. The first inequality $x + 2y \leq 2$ represents the region on or below the line $x + 2y = 2$.
The second inequality $x + 2y \geq 4$ represents the region on or above the line $x + 2y = 4$.
Since there is no point $(x, y)$ that satisfies both $x + 2y \leq 2$ and $x + 2y \geq 4$ simultaneously,there is no common region that satisfies all the constraints.
Therefore,the problem has no feasible region.

(A) To find the maximum value of the objective function $z = 2x + 3y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At point $A(3, 3): z = 2(3) + 3(3) = 6 + 9 = 15$
$2$. At point $B(20, 3): z = 2(20) + 3(3) = 40 + 9 = 49$
$3$. At point $C(20, 10): z = 2(20) + 3(10) = 40 + 30 = 70$
$4$. At point $D(18, 12): z = 2(18) + 3(12) = 36 + 36 = 72$
$5$. At point $E(12, 12): z = 2(12) + 3(12) = 24 + 36 = 60$
Comparing these values,the maximum value of $z$ is $72$.

(B) To find the minimum value of the objective function $z = 2x + 3y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At $A(3, 3): z = 2(3) + 3(3) = 6 + 9 = 15$
$2$. At $B(20, 3): z = 2(20) + 3(3) = 40 + 9 = 49$
$3$. At $C(20, 10): z = 2(20) + 3(10) = 40 + 30 = 70$
$4$. At $D(18, 12): z = 2(18) + 3(12) = 36 + 36 = 72$
$5$. At $E(12, 12): z = 2(12) + 3(12) = 24 + 36 = 60$
Comparing these values,the minimum value of $z$ is $15$ at point $A(3, 3)$.

(B) To find the minimum value of the objective function $z = 5x + 4y$,we evaluate $z$ at the corner points of the feasible region.
From the graph,the corner points of the feasible region are $(0, 40)$,$(5, 30)$,$(20, 15)$,and $(50, 0)$.
Evaluating $z$ at each point:
$1$. At $(0, 40)$: $z = 5(0) + 4(40) = 0 + 160 = 160$
$2$. At $(5, 30)$: $z = 5(5) + 4(30) = 25 + 120 = 145$
$3$. At $(20, 15)$: $z = 5(20) + 4(15) = 100 + 60 = 160$
$4$. At $(50, 0)$: $z = 5(50) + 4(0) = 250 + 0 = 250$
Comparing these values,the minimum value is $145$ at the point $(5, 30)$.

(B) To find the minimum and maximum values of the objective function $z = 8x + 12y$,we evaluate $z$ at each corner point:
$1. \text{At } (0,4): z = 8(0) + 12(4) = 48$
$2. \text{At } (6,0): z = 8(6) + 12(0) = 48$
$3. \text{At } (12,0): z = 8(12) + 12(0) = 96$
$4. \text{At } (12,16): z = 8(12) + 12(16) = 96 + 192 = 288$
$5. \text{At } (0,10): z = 8(0) + 12(10) = 120$
Comparing these values:
- The minimum value is $48$,which occurs at both $(0,4)$ and $(6,0)$.
- The maximum value is $288$,which occurs at $(12,16)$.
Matching the given requirements:
$(i)$ Minimum value occurs at $(0,4)$ (or $(6,0)$).
$(ii)$ Maximum value occurs at $(12,16)$.
$(iii)$ Maximum of $z$ is $288$.
$(iv)$ Minimum of $z$ is $48$.
Thus,option $(B)$ is the correct match.

(D) To find the maximum and minimum values of $Z = 22x + 18y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0,0): Z = 22(0) + 18(0) = 0$
$2$. At $(16,0): Z = 22(16) + 18(0) = 352$
$3$. At $(8,12): Z = 22(8) + 18(12) = 176 + 216 = 392$
$4$. At $(0,20): Z = 22(0) + 18(20) = 360$
Comparing these values,the maximum value $m = 392$ and the minimum value $n = 0$.
Therefore,$m + n = 392 + 0 = 392$.

(C) To find the corner points of the feasible region,we examine the intersection of the lines $x+2y = 10$ and $3x+4y = 24$ with the axes and each other.
$1$. For $(0, 6)$:
$x+2y = 0 + 12 = 12 \geq 10$ (True)
$3x+4y = 0 + 24 = 24 \leq 24$ (True)
Since it satisfies all constraints,$(0, 6)$ is a corner point.
$2$. For $(4, 3)$:
$x+2y = 4 + 6 = 10 \geq 10$ (True)
$3x+4y = 12 + 12 = 24 \leq 24$ (True)
Since it satisfies all constraints,$(4, 3)$ is a corner point.
$3$. For $(3, 4)$:
$x+2y = 3 + 8 = 11 \geq 10$ (True)
$3x+4y = 9 + 16 = 25 \not\leq 24$ (False)
Since it violates the constraint $3x+4y \leq 24$,$(3, 4)$ is not in the feasible region.
$4$. For $(0, 5)$:
$x+2y = 0 + 10 = 10 \geq 10$ (True)
$3x+4y = 0 + 20 = 20 \leq 24$ (True)
Since it satisfies all constraints,$(0, 5)$ is a corner point.
Therefore,$(3, 4)$ is not a corner point of the feasible region.

(B) The constraints are $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$.
$1$. Since $x \geq 6$,it automatically implies $x \geq 0$. Thus,$x \geq 0$ is a redundant constraint.
$2$. Since $y \geq 2$,it automatically implies $y \geq 0$. Thus,$y \geq 0$ is a redundant constraint.
$3$. For the constraint $2x + y \geq 10$,we substitute the minimum values $x = 6$ and $y = 2$: $2(6) + 2 = 12 + 2 = 14$. Since $14 \geq 10$,the constraint $2x + y \geq 10$ is satisfied by the region defined by $x \geq 6$ and $y \geq 2$. Thus,$2x + y \geq 10$ is also a redundant constraint.
Therefore,the redundant constraints are $2x + y \geq 10, x \geq 0, y \geq 0$.

(A) To find the maximum value of the objective function $Z = 2x + 3y$,we evaluate $Z$ at each corner point of the feasible region:

Corner Point $(x, y)$	Value of $Z = 2x + 3y$
$A(3, 3)$	$Z = 2(3) + 3(3) = 6 + 9 = 15$
$B(20, 3)$	$Z = 2(20) + 3(3) = 40 + 9 = 49$
$C(20, 10)$	$Z = 2(20) + 3(10) = 40 + 30 = 70$
$D(18, 12)$	$Z = 2(18) + 3(12) = 36 + 36 = 72$
$E(12, 12)$	$Z = 2(12) + 3(12) = 24 + 36 = 60$

Comparing the values of $Z$ at all corner points,the maximum value is $72$ at point $D(18, 12)$.

(B) The given line passes through the points $B\left(-\frac{3}{4}, 0\right)$ and $A\left(0, \frac{3}{2}\right)$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,where $a = -\frac{3}{4}$ and $b = \frac{3}{2}$.
Substituting the values,we get $\frac{x}{-\frac{3}{4}} + \frac{y}{\frac{3}{2}} = 1$.
This simplifies to $-\frac{4x}{3} + \frac{2y}{3} = 1$.
Multiplying by $3$,we get $-4x + 2y = 3$,which can be rewritten as $4x - 2y = -3$.
The shaded region does not contain the origin $(0,0)$.
Testing the origin $(0,0)$ in the inequality $4x - 2y \leq -3$,we get $4(0) - 2(0) \leq -3$,which means $0 \leq -3$. This is false.
Since the origin does not satisfy the inequality,the shaded region represents the half-plane $4x - 2y \leq -3$.

(B) The feasible region is determined by the constraints $x + 2y \leq 2$,$x \geq 0$,and $y \geq 0$.
The corner points of the feasible region are $(0, 0)$,$(2, 0)$,and $(0, 1)$.
We evaluate $Z = 3x + 2y$ at these corner points:
At $(0, 0)$: $Z = 3(0) + 2(0) = 0$.
At $(2, 0)$: $Z = 3(2) + 2(0) = 6$.
At $(0, 1)$: $Z = 3(0) + 2(1) = 2$.
The maximum value of $Z$ is $6$,which occurs at the point $(2, 0)$.

(D) The feasible region is determined by the constraints $2x+y \leq 20$,$x+2y \leq 20$,$x \geq 0$,and $y \geq 0$.
The corner points of the feasible region are $O(0,0)$,$A(10,0)$,$B(\frac{20}{3}, \frac{20}{3})$,and $C(0,10)$.
We evaluate $Z = x+3y$ at each corner point:
At $O(0,0)$,$Z = 0 + 3(0) = 0$.
At $A(10,0)$,$Z = 10 + 3(0) = 10$.
At $B(\frac{20}{3}, \frac{20}{3})$,$Z = \frac{20}{3} + 3(\frac{20}{3}) = \frac{20}{3} + 20 = \frac{80}{3} \approx 26.67$.
At $C(0,10)$,$Z = 0 + 3(10) = 30$.
Comparing these values,the maximum value of $Z$ is $30$ at point $C(0,10)$.

(C) To find which point lies in the solution set,we test each point against all given constraints:
$1$. For $(2, 3)$: $x + 2y = 2 + 2(3) = 8$. Since $8 \geq 11$ is false,$(2, 3)$ is not in the solution set.
$2$. For $(3, 2)$: $x + 2y = 3 + 2(2) = 7$. Since $7 \geq 11$ is false,$(3, 2)$ is not in the solution set.
$3$. For $(3, 4)$:
$x + 2y = 3 + 2(4) = 11$. Since $11 \geq 11$ is true.
$3x + 4y = 3(3) + 4(4) = 9 + 16 = 25$. Since $25 \leq 30$ is true.
$2x + 5y = 2(3) + 5(4) = 6 + 20 = 26$. Since $26 \leq 30$ is true.
Since all constraints are satisfied,$(3, 4)$ is in the solution set.
$4$. For $(4, 3)$: $x + 2y = 4 + 2(3) = 10$. Since $10 \geq 11$ is false,$(4, 3)$ is not in the solution set.

(C) To identify the redundant constraint,we analyze the feasible region defined by the given inequalities:
$1$) $2x - y \leq 8$
$2$) $x + y \leq 20$
$3$) $-x + y \geq -10$ (which is $y \geq x - 10$)
$4$) $x \geq 0$
$5$) $y \geq 0$
Consider the constraints $x \geq 0$ and $y \geq 0$. If $x \geq 0$,then $x - 10 \geq -10$. Since $y \geq 0$,the condition $y \geq x - 10$ is automatically satisfied for all $x \geq 0$ and $y \geq 0$ because $y$ is non-negative and $x - 10$ is at most $-10$ when $x=0$. Thus,the constraint $-x + y \geq -10$ does not restrict the feasible region further than the non-negativity constraints $x \geq 0$ and $y \geq 0$. Therefore,$-x + y \geq -10$ is a redundant constraint.

(A) To find the minimum value of the objective function $z = 30x - 30y + 1800$,we evaluate $z$ at each corner point of the feasible region:

Corner Point $(x, y)$	Value of $z = 30x - 30y + 1800$
$(15, 0)$	$30(15) - 30(0) + 1800 = 450 + 1800 = 2250$
$(15, 15)$	$30(15) - 30(15) + 1800 = 450 - 450 + 1800 = 1800$
$(10, 20)$	$30(10) - 30(20) + 1800 = 300 - 600 + 1800 = 1500$
$(0, 20)$	$30(0) - 30(20) + 1800 = 0 - 600 + 1800 = 1200$
$(0, 15)$	$30(0) - 30(15) + 1800 = 0 - 450 + 1800 = 1350$

Comparing the values,the minimum value of $z$ is $1200$,which occurs at the point $(0, 20)$.