Word problem of Linear programming Questions in English

(A) The shaded region in the figure is the feasible region determined by the system of constraints $(2)$ to $(4)$. We observe that the feasible region $OABC$ is bounded. So,we now use the Corner Point Method to determine the maximum value of $Z$.
The coordinates of the corner points $O, A, B$ and $C$ are $(0, 0), (30, 0), (20, 30)$ and $(0, 50)$ respectively. Now we evaluate $Z$ at each corner point.

Corner Point	Corresponding value of $Z = 4x + y$
$(0, 0)$	$4(0) + 0 = 0$
$(30, 0)$	$4(30) + 0 = 120$
$(20, 30)$	$4(20) + 30 = 110$
$(0, 50)$	$4(0) + 50 = 50$

Hence,the maximum value of $Z$ is $120$ at the point $(30, 0)$.

(B) Let the mixture contain $x$ $kg$ of Food $'I'$ and $y$ $kg$ of Food $'II'$. Clearly,$x \geq 0, y \geq 0$.
We make the following table from the given data:

Resources	Food $I$ $(x)$	Food $II$ $(y)$	Minimum Requirement
Vitamin $A$ (units/kg)	$2$	$1$	$8$
Vitamin $C$ (units/kg)	$1$	$2$	$10$
Cost (Rs/kg)	$50$	$70$	Minimize $Z$

Since the mixture must contain at least $8$ units of vitamin $A$ and $10$ units of vitamin $C$,we have the constraints:
$2x + y \geq 8$
$x + 2y \geq 10$
Total cost $Z$ of purchasing $x$ $kg$ of food $'I'$ and $y$ $kg$ of Food $'II'$ is $Z = 50x + 70y$.
Hence,the mathematical formulation of the problem is:
Minimize $Z = 50x + 70y$ subject to the constraints:
$2x + y \geq 8$
$x + 2y \geq 10$
$x, y \geq 0$
Let us graph the inequalities. The feasible region is unbounded. Let us evaluate $Z$ at the corner points $A(0, 8)$,$B(2, 4)$,and $C(10, 0)$.

Corner point	$Z = 50x + 70y$
$(0, 8)$	$560$
$(2, 4)$	$380$ (Minimum)
$(10, 0)$	$500$

The smallest value of $Z$ is $380$ at the point $(2, 4)$. Since the feasible region is unbounded,we check the inequality $50x + 70y < 380$ or $5x + 7y < 38$. As this region has no common points with the feasible region,the minimum value is indeed $380$.
Thus,the optimal mixing strategy is to mix $2$ $kg$ of Food $'I'$ and $4$ $kg$ of Food $'II'$,with a minimum cost of Rs $380$.

(A) Let $x$ hectares of land be allocated to crop $X$ and $y$ hectares to crop $Y$.
Obviously,$x \geq 0, y \geq 0.$
Profit per hectare on crop $X = \text{Rs } 10,500$
Profit per hectare on crop $Y = \text{Rs } 9,000$
Therefore,total profit $Z = 10,500x + 9,000y$
The mathematical formulation of the problem is as follows:
Maximise $Z = 10,500x + 9,000y$
Subject to the constraints:
$x + y \leq 50$ (constraint related to land) $... (1)$
$20x + 10y \leq 800$ (constraint related to use of herbicide)
i.e.,$2x + y \leq 80$ $... (2)$
$x \geq 0, y \geq 0$ (non-negative constraint) $... (3)$
The feasible region $OABC$ is bounded by the vertices $O(0,0), A(40,0), B(30,20),$ and $C(0,50)$.
Evaluating the objective function $Z = 10,500x + 9,000y$ at these vertices:

Corner Point	$Z = 10,500x + 9,000y$
$O(0,0)$	$0$
$A(40,0)$	$4,20,000$
$B(30,20)$	$10,500(30) + 9,000(20) = 3,15,000 + 1,80,000 = 4,95,000$ (Maximum)
$C(0,50)$	$4,50,000$

Hence,the society will get the maximum profit of Rs $4,95,000$ by allocating $30$ hectares for crop $X$ and $20$ hectares for crop $Y$.

(A) Let $x$ be the number of pieces of Model $A$ and $y$ be the number of pieces of Model $B$.
The objective function is to maximize profit $Z = 8000x + 12000y$.
The constraints based on labour hours are:
$9x + 12y \leq 180 \implies 3x + 4y \leq 60$ (Fabricating)
$x + 3y \leq 30$ (Finishing)
$x \geq 0, y \geq 0$ (Non-negativity)
The corner points of the feasible region are $O(0,0)$,$A(20,0)$,$B(12,6)$,and $C(0,10)$.
Evaluating $Z$ at corner points:

Corner Point	$Z = 8000x + 12000y$
$O(0,0)$	$0$
$A(20,0)$	$8000(20) + 0 = 1,60,000$
$B(12,6)$	$8000(12) + 12000(6) = 96000 + 72000 = 1,68,000$
$C(0,10)$	$8000(0) + 12000(10) = 1,20,000$

The maximum profit is Rs $1,68,000$ at $x=12$ and $y=6$.

(A) Let the mixture contain $x$ kg of food $P$ and $y$ kg of food $Q$. Therefore,$x \geq 0$ and $y \geq 0$. The given information is compiled in the table below:

Food	Vitamin $A$ (units/kg)	Vitamin $B$ (units/kg)	Cost (Rs/kg)
$P$	$3$	$5$	$60$
$Q$	$4$	$2$	$80$
Requirement	$8$	$11$	-

The constraints are:
$3x + 4y \geq 8$
$5x + 2y \geq 11$
$x, y \geq 0$
Objective function to minimize: $Z = 60x + 80y$.
The corner points of the feasible region are $A(\frac{8}{3}, 0)$,$B(2, \frac{1}{2})$,and $C(0, \frac{11}{2})$.
Evaluating $Z$ at corner points:

Corner Point	$Z = 60x + 80y$
$A(\frac{8}{3}, 0)$	$60(\frac{8}{3}) + 80(0) = 160$
$B(2, \frac{1}{2})$	$60(2) + 80(\frac{1}{2}) = 120 + 40 = 160$
$C(0, \frac{11}{2})$	$60(0) + 80(\frac{11}{2}) = 440$

Since the feasible region is unbounded,we check if $60x + 80y < 160$ has any common points with the feasible region. The line $3x + 4y < 8$ does not intersect the feasible region. Thus,the minimum cost is Rs $160$ at any point on the line segment joining $A$ and $B$.

(30) Let $x$ be the number of cakes of the first kind and $y$ be the number of cakes of the second kind.
Therefore,$x \geq 0$ and $y \geq 0$.
The given information can be summarized in the following table:

Type of Cake	Flour $(g)$	Fat $(g)$
First kind $(x)$	$200$	$25$
Second kind $(y)$	$100$	$50$
Availability	$5000$	$1000$

Constraints:
$200x + 100y \leq 5000 \Rightarrow 2x + y \leq 50$
$25x + 50y \leq 1000 \Rightarrow x + 2y \leq 40$
Objective function: Maximize $Z = x + y$.
The feasible region is determined by the constraints $2x + y \leq 50$,$x + 2y \leq 40$,$x \geq 0$,and $y \geq 0$. The corner points of the feasible region are $O(0, 0)$,$A(25, 0)$,$B(20, 10)$,and $C(0, 20)$.
Evaluating $Z = x + y$ at corner points:

Corner Point	$Z = x + y$
$O(0, 0)$	$0$
$A(25, 0)$	$25$
$B(20, 10)$	$30$ (Maximum)
$C(0, 20)$	$20$

Thus,the maximum number of cakes that can be made is $30$ ($20$ of the first kind and $10$ of the second kind).

(A) Let the number of tennis rackets be $x$ and the number of cricket bats be $y$.
The machine time constraint is $1.5x + 3y \leq 42$.
The craftsman's time constraint is $3x + y \leq 24$.
Since the factory works at full capacity, we equate these to the maximum available time:
$1.5x + 3y = 42$ --- $(1)$
$3x + y = 24$ --- $(2)$
From equation $(2)$, we get $y = 24 - 3x$.
Substitute this into equation $(1)$:
$1.5x + 3(24 - 3x) = 42$
$1.5x + 72 - 9x = 42$
$-7.5x = 42 - 72$
$-7.5x = -30$
$x = \frac{30}{7.5} = 4$
Now, substitute $x = 4$ into $y = 24 - 3x$:
$y = 24 - 3(4) = 24 - 12 = 12$.
Therefore, the factory must make $4$ tennis rackets and $12$ cricket bats.

(A) The given information can be compiled in a table as follows:

Item	Tennis Racket $(x)$	Cricket Bat $(y)$	Availability
Machine Time $(h)$	$1.5$	$3$	$42$
Craftsman's Time $(h)$	$3$	$1$	$24$

Let $x$ be the number of tennis rackets and $y$ be the number of cricket bats produced.
The constraints are:
$1.5x + 3y \leq 42$
$3x + y \leq 24$
$x, y \geq 0$
The objective function is to maximize profit $Z = 20x + 10y$.
Solving the intersection of lines $1.5x + 3y = 42$ and $3x + y = 24$:
From $3x + y = 24$,we get $y = 24 - 3x$.
Substituting into $1.5x + 3(24 - 3x) = 42$:
$1.5x + 72 - 9x = 42$
$-7.5x = -30$
$x = 4$
Then $y = 24 - 3(4) = 12$.
The corner points of the feasible region are $(0, 0), (8, 0), (4, 12), (0, 14)$.
Evaluating $Z = 20x + 10y$ at these points:
- At $(0, 0): Z = 0$
- At $(8, 0): Z = 20(8) + 10(0) = 160$
- At $(4, 12): Z = 20(4) + 10(12) = 80 + 120 = 200$
- At $(0, 14): Z = 20(0) + 10(14) = 140$
The maximum profit is $Rs. 200$.

(B) Let the manufacturer produce $x$ packages of nuts and $y$ packages of bolts.
Therefore,$x \geq 0$ and $y \geq 0$.
The given information can be compiled in a table as follows:

Machine	Nuts	Bolts	Availability
Machine $A$ $(h)$	$1$	$3$	$12$
Machine $B$ $(h)$	$3$	$1$	$12$

The profit on a package of nuts is $Rs.\,17.50$ and on a package of bolts is $Rs.\,7$. Therefore,the constraints are:
$x + 3y \leq 12$
$3x + y \leq 12$
Total profit,$Z = 17.5x + 7y$.
The mathematical formulation of the given problem is:
Maximise $Z = 17.5x + 7y$ subject to the constraints:
$x + 3y \leq 12$
$3x + y \leq 12$
$x, y \geq 0$
The feasible region determined by the system of constraints has corner points $O(0,0)$,$A(4,0)$,$B(3,3)$,and $C(0,4)$.
The values of $Z$ at these corner points are as follows:

Corner point	$Z = 17.5x + 7y$
$O(0,0)$	$0$
$A(4,0)$	$70$
$B(3,3)$	$73.5$ (Maximum)
$C(0,4)$	$28$

The maximum value of $Z$ is $Rs.\,73.50$ at $(3,3)$.
Thus,$3$ packages of nuts and $3$ packages of bolts should be produced each day to get the maximum profit of $Rs.\,73.50$.

(A) Let the factory manufacture $x$ packages of screws $A$ and $y$ packages of screws $B$ each day.
The given information can be summarized in the following table:

Machine	Screw $A$ (min)	Screw $B$ (min)	Availability (min)
Automatic	$4$	$6$	$240$
Hand-operated	$6$	$3$	$240$

The objective is to maximize profit $Z = 7x + 10y$.
Subject to constraints:
$4x + 6y \leq 240 \implies 2x + 3y \leq 120$
$6x + 3y \leq 240 \implies 2x + y \leq 80$
$x, y \geq 0$
The corner points of the feasible region are $O(0,0)$, $A(40,0)$, $B(30,20)$, and $C(0,40)$.
Evaluating $Z$ at corner points:

Corner Point	$Z = 7x + 10y$
$O(0,0)$	$0$
$A(40,0)$	$7(40) + 10(0) = 280$
$B(30,20)$	$7(30) + 10(20) = 210 + 200 = 410$
$C(0,40)$	$7(0) + 10(40) = 400$

The maximum profit is $Rs. \, 410$ at $x = 30$ and $y = 20$.

(A) Let the cottage industry manufacture $x$ pedestal lamps and $y$ wooden shades.
Therefore,$x \geq 0$ and $y \geq 0$.
The given information can be compiled in a table as follows:

Item	Lamps $(x)$	Shades $(y)$	Availability
Grinding/cutting machine $(h)$	$2$	$1$	$12$
Sprayer $(h)$	$3$	$2$	$20$

The profit on a lamp is $Rs. 5$ and on a shade is $Rs. 3$.
Therefore,the constraints are:
$2x + y \leq 12$
$3x + 2y \leq 20$
Total profit,$Z = 5x + 3y$.
The mathematical formulation of the problem is:
Maximize $Z = 5x + 3y$ subject to:
$2x + y \leq 12$
$3x + 2y \leq 20$
$x, y \geq 0$
The feasible region is determined by the intersection of these constraints. The corner points of the feasible region are $O(0,0)$,$A(6,0)$,$B(4,4)$,and $C(0,10)$.
The values of $Z$ at these corner points are:

Corner point	$Z = 5x + 3y$
$O(0,0)$	$0$
$A(6,0)$	$5(6) + 3(0) = 30$
$B(4,4)$	$5(4) + 3(4) = 20 + 12 = 32$
$C(0,10)$	$5(0) + 3(10) = 30$

The maximum value of $Z$ is $32$ at point $B(4,4)$.
Thus,the manufacturer should produce $4$ pedestal lamps and $4$ wooden shades to maximize his profit.

(A) Let the company manufacture $x$ souvenirs of type $A$ and $y$ souvenirs of type $B$.
The given information can be compiled in a table as follows:

Process	Type $A$	Type $B$	Availability
Cutting (min)	$5$	$8$	$200$
Assembling (min)	$10$	$8$	$240$

The constraints are:
$5x + 8y \leq 200$
$10x + 8y \leq 240 \implies 5x + 4y \leq 120$
$x, y \geq 0$
Objective function to maximize profit $Z = 5x + 6y$.
The feasible region is bounded by the corner points $O(0,0)$,$A(24,0)$,$B(8,20)$,and $C(0,25)$.
Evaluating $Z$ at corner points:

Corner Point	$Z = 5x + 6y$
$A(24,0)$	$5(24) + 6(0) = 120$
$B(8,20)$	$5(8) + 6(20) = 40 + 120 = 160$
$C(0,25)$	$5(0) + 6(25) = 150$

The maximum profit is $Rs. 160$ at $(8, 20)$. Thus,the company should manufacture $8$ souvenirs of type $A$ and $20$ souvenirs of type $B$.

(A) Let the merchant stock $x$ desktop models and $y$ portable models.
The cost of a desktop model is $Rs.\,25000$ and of a portable model is $Rs.\,40000$.
Given that the merchant can invest a maximum of $Rs.\,70$ lakhs $(Rs.\,70,00,000)$:
$25000x + 40000y \leq 7000000$
Dividing by $5000$,we get:
$5x + 8y \leq 1400$ ... $(1)$
The total monthly demand of computers will not exceed $250$ units:
$x + y \leq 250$ ... $(2)$
Also,$x \geq 0$ and $y \geq 0$ ... $(3)$
The profit function to be maximized is:
$Z = 4500x + 5000y$
We find the corner points of the feasible region by solving the equations $5x + 8y = 1400$ and $x + y = 250$:
From $x + y = 250$,$y = 250 - x$. Substituting in $(1)$:
$5x + 8(250 - x) = 1400$
$5x + 2000 - 8x = 1400$
$-3x = -600 \implies x = 200$
$y = 250 - 200 = 50$
The corner points of the feasible region are $(0, 0)$,$(250, 0)$,$(200, 50)$,and $(0, 175)$.
Evaluating $Z$ at these points:

Corner Point	$Z = 4500x + 5000y$
$(0, 0)$	$0$
$(250, 0)$	$4500(250) = 1125000$
$(200, 50)$	$4500(200) + 5000(50) = 900000 + 250000 = 1150000$
$(0, 175)$	$5000(175) = 875000$

The maximum profit is $Rs.\,1150000$ at $(200, 50)$.
Thus,the merchant should stock $200$ desktop models and $50$ portable models.

(A) Let the diet contain $x$ units of food $F_{1}$ and $y$ units of food $F_{2}$.
The given information can be compiled in a table as follows:

Food	Vitamin $A$ (units)	Minerals (units)	Cost (Rs)
$F_{1} (x)$	$3$	$4$	$4$
$F_{2} (y)$	$6$	$3$	$6$
Requirement	$80$	$100$	-

The mathematical formulation is:
Minimize $Z = 4x + 6y$ subject to:
$3x + 6y \geq 80$
$4x + 3y \geq 100$
$x, y \geq 0$
The corner points of the feasible region are $A(\frac{80}{3}, 0)$,$B(24, \frac{4}{3})$,and $C(0, \frac{100}{3})$.
Evaluating $Z$ at corner points:
- At $A(\frac{80}{3}, 0)$,$Z = 4(\frac{80}{3}) + 6(0) = \frac{320}{3} \approx 106.67$
- At $B(24, \frac{4}{3})$,$Z = 4(24) + 6(\frac{4}{3}) = 96 + 8 = 104$
- At $C(0, \frac{100}{3})$,$Z = 4(0) + 6(\frac{100}{3}) = 200$
Since the feasible region is unbounded,we check if $4x + 6y < 104$ has any common points with the feasible region. The line $4x + 6y = 104$ (or $2x + 3y = 52$) does not intersect the feasible region. Thus,the minimum cost is $Rs. 104$.

(A) Let the farmer buy $x\,kg$ of fertilizer $F_{1}$ and $y\,kg$ of fertilizer $F_{2}$.
Constraints:
$1$. Nitrogen requirement: $0.10x + 0.05y \geq 14 \implies 10x + 5y \geq 1400 \implies 2x + y \geq 280$.
$2$. Phosphoric acid requirement: $0.06x + 0.10y \geq 14 \implies 6x + 10y \geq 1400 \implies 3x + 5y \geq 700$.
$3$. Non-negativity: $x \geq 0, y \geq 0$.
Objective function to minimize: $Z = 6x + 5y$.
Finding corner points of the feasible region:
- Intersection of $2x + y = 280$ and $3x + 5y = 700$:
Multiply first by $5$: $10x + 5y = 1400$.
Subtract second: $(10x - 3x) = 1400 - 700 \implies 7x = 700 \implies x = 100$.
Substitute $x=100$ in $2x + y = 280 \implies 200 + y = 280 \implies y = 80$.
Corner point $B(100, 80)$.
- Intersection with axes:
For $2x + y = 280$,intercepts are $(140, 0)$ and $(0, 280)$.
For $3x + 5y = 700$,intercepts are $(700/3, 0)$ and $(0, 140)$.
The feasible region is unbounded with corner points $A(700/3, 0)$,$B(100, 80)$,and $C(0, 280)$.
Evaluating $Z$ at corner points:
- $Z(A) = 6(700/3) + 5(0) = 1400$.
- $Z(B) = 6(100) + 5(80) = 600 + 400 = 1000$.
- $Z(C) = 6(0) + 5(280) = 1400$.
Since the region is unbounded,we check $6x + 5y < 1000$. The line $6x + 5y = 1000$ does not intersect the interior of the feasible region. Thus,the minimum cost is $Rs\,1000$ at $x=100, y=80$.

(A) Let $x$ and $y$ be the number of packets of food $P$ and $Q$ respectively. The objective is to minimise $Z = 6x + 3y$ (vitamin $A$).
Constraints are:
$12x + 3y \geq 240 \implies 4x + y \geq 80$
$4x + 20y \geq 460 \implies x + 5y \geq 115$
$6x + 4y \leq 300 \implies 3x + 2y \leq 150$
$x, y \geq 0$
The feasible region is bounded by the corner points $L(2, 72)$,$M(15, 20)$,and $N(40, 15)$.

Corner Point	$Z = 6x + 3y$
$L(2, 72)$	$6(2) + 3(72) = 12 + 216 = 228$
$M(15, 20)$	$6(15) + 3(20) = 90 + 60 = 150$
$N(40, 15)$	$6(40) + 3(15) = 240 + 45 = 285$

The minimum value of $Z$ is $150$ at point $(15, 20)$. Thus,$15$ packets of food $P$ and $20$ packets of food $Q$ should be used.

(A) Let $x$ and $y$ be the number of items $M$ and $N$ produced respectively.
Total profit $Z = 600x + 400y$.
Constraints:
$x + 2y \leq 12$ (Machine $I$)
$2x + y \leq 12$ (Machine $II$)
$x + 1.25y \geq 5$ (Machine $III$)
$x, y \geq 0$
Solving the system of inequalities,the feasible region is bounded by the vertices $(5, 0), (6, 0), (4, 4), (0, 6), (0, 4)$.
Evaluating $Z$ at corner points:
At $(5, 0): Z = 600(5) + 400(0) = 3000$
At $(6, 0): Z = 600(6) + 400(0) = 3600$
At $(4, 4): Z = 600(4) + 400(4) = 2400 + 1600 = 4000$
At $(0, 6): Z = 600(0) + 400(6) = 2400$
At $(0, 4): Z = 600(0) + 400(4) = 1600$
The maximum profit is $Rs. \, 4000$ at $x = 4, y = 4$.

(A) Let $x$ and $y$ be the units transported from factory $P$ to depots $A$ and $B$ respectively. Then,$8-x-y$ units are transported from $P$ to $C$.
The units transported from $Q$ are:
To $A$: $5-x$
To $B$: $5-y$
To $C$: $6-(5-x)-(5-y) = x+y-4$
The constraints are:
$x \geq 0, y \geq 0$
$x \leq 5, y \leq 5$
$x+y \leq 8$
$x+y \geq 4$
The total cost function $Z$ is:
$Z = 160x + 100y + 150(8-x-y) + 100(5-x) + 120(5-y) + 100(x+y-4)$
$Z = 160x + 100y + 1200 - 150x - 150y + 500 - 100x + 600 - 120y + 100x + 100y - 400$
$Z = 10x - 70y + 1900 = 10(x - 7y + 190)$
Evaluating $Z$ at corner points of the feasible region:
$(0,4): Z = 10(0 - 28 + 190) = 1620$
$(0,5): Z = 10(0 - 35 + 190) = 1550$
$(3,5): Z = 10(3 - 35 + 190) = 1580$
$(5,3): Z = 10(5 - 21 + 190) = 1740$
$(5,0): Z = 10(5 - 0 + 190) = 1950$
$(4,0): Z = 10(4 - 0 + 190) = 1940$
The minimum cost is $1550$ at $(0,5)$.

(A) Let the diet contain $x$ and $y$ packets of foods $P$ and $Q$ respectively. The objective is to maximize $Z = 6x + 3y$. The constraints are:
$12x + 3y \geq 240 \implies 4x + y \geq 80$
$4x + 20y \geq 460 \implies x + 5y \geq 115$
$6x + 4y \leq 300 \implies 3x + 2y \leq 150$
$x, y \geq 0$.
Solving the intersection points of the lines:
$4x + y = 80$ and $x + 5y = 115$ gives $A(15, 20)$.
$4x + y = 80$ and $3x + 2y = 150$ gives $B(2, 72)$.
$x + 5y = 115$ and $3x + 2y = 150$ gives $C(40, 15)$.
Evaluating $Z = 6x + 3y$ at corner points:

Corner Point	$Z = 6x + 3y$
$A(15, 20)$	$6(15) + 3(20) = 150$
$B(2, 72)$	$6(2) + 3(72) = 228$
$C(40, 15)$	$6(40) + 3(15) = 285$

The maximum value is $285$ at $(40, 15)$. Thus,$40$ packets of $P$ and $15$ packets of $Q$ are required.

(A) Let the farmer mix $x$ bags of brand $P$ and $y$ bags of brand $Q$.
The objective is to minimize the cost $Z = 250x + 200y$.
Subject to the constraints:
$3x + 1.5y \geq 18$ (Nutrient $A$)
$2.5x + 11.25y \geq 45$ (Nutrient $B$)
$2x + 3y \geq 24$ (Nutrient $C$)
$x, y \geq 0$
Solving the intersection points of the lines:
$1$. $3x + 1.5y = 18$ and $2x + 3y = 24$ intersect at $(3, 6)$.
$2$. $2.5x + 11.25y = 45$ and $2x + 3y = 24$ intersect at $(9, 2)$.
$3$. The boundary lines intersect the axes at $(18, 0)$ and $(0, 12)$.
The corner points of the feasible region are $(18, 0), (9, 2), (3, 6), (0, 12)$.
Evaluating $Z = 250x + 200y$ at these points:
- At $(18, 0): Z = 250(18) + 200(0) = 4500$
- At $(9, 2): Z = 250(9) + 200(2) = 2250 + 400 = 2650$
- At $(3, 6): Z = 250(3) + 200(6) = 750 + 1200 = 1950$
- At $(0, 12): Z = 250(0) + 200(12) = 2400$
Since the feasible region is unbounded,we check if $250x + 200y < 1950$ has any common points with the feasible region. The line $250x + 200y = 1950$ (or $5x + 4y = 39$) does not intersect the feasible region except at $(3, 6)$. Thus,the minimum cost is $Rs. 1950$ at $x=3, y=6$.

(A) Let the mixture contain $x$ $kg$ of food $X$ and $y$ $kg$ of food $Y$.
The mathematical formulation of the given problem is as follows:
Minimize $Z = 16x + 20y$ ... $(1)$
Subject to the constraints:
$x + 2y \geq 10$ ... $(2)$
$2x + 2y \geq 12$ (which simplifies to $x + y \geq 6$) ... $(3)$
$3x + y \geq 8$ ... $(4)$
$x, y \geq 0$ ... $(5)$
The feasible region determined by the system of constraints is unbounded.
The corner points of the feasible region are $A(10, 0)$,$B(2, 4)$,$C(1, 5)$,and $D(0, 8)$.
The values of $Z$ at these corner points are as follows:

Corner point	$Z = 16x + 20y$
$A(10, 0)$	$160$
$B(2, 4)$	$112$
$C(1, 5)$	$116$
$D(0, 8)$	$160$

As the feasible region is unbounded,we check if $16x + 20y < 112$ has any common points with the feasible region.
Since the line $16x + 20y = 112$ does not pass through the interior of the feasible region,the minimum value is $112$ at $(2, 4)$.
Thus,the least cost of the mixture is $Rs. 112$.

(A) Let $x$ and $y$ be the number of toys of type $A$ and type $B$ manufactured in a day,respectively.
The constraints based on machine time (in minutes) are:
Machine $I$: $12x + 6y \leq 360 \implies 2x + y \leq 60$
Machine $II$: $18x + 0y \leq 360 \implies x \leq 20$
Machine $III$: $6x + 9y \leq 360 \implies 2x + 3y \leq 120$
Non-negativity: $x, y \geq 0$
Objective function to maximize profit: $Z = 7.5x + 5y$
The feasible region is bounded by the vertices $(0, 0), (20, 0), (20, 20), (15, 30), (0, 40)$.
Evaluating $Z$ at corner points:
$1$. At $(0, 0): Z = 0$
$2$. At $(20, 0): Z = 7.5(20) + 5(0) = 150$
$3$. At $(20, 20): Z = 7.5(20) + 5(20) = 150 + 100 = 250$
$4$. At $(15, 30): Z = 7.5(15) + 5(30) = 112.5 + 150 = 262.5$
$5$. At $(0, 40): Z = 7.5(0) + 5(40) = 200$
The maximum profit is $Rs. \, 262.5$ at $(15, 30)$. Thus,$15$ toys of type $A$ and $30$ toys of type $B$ should be manufactured.

(D) Let godown $A$ supply $x$ and $y$ quintals of grain to the shops $D$ and $E$ respectively. Then,$(100-x-y)$ will be supplied to shop $F$.
The requirement at shop $D$ is $60$ quintals. Since $x$ quintals are transported from godown $A$,the remaining $(60-x)$ quintals will be transported from godown $B$.
Similarly,$(50-y)$ quintals and $40-(100-x-y) = (x+y-60)$ quintals will be transported from godown $B$ to shop $E$ and $F$ respectively.
Total transportation cost $z$ is given by:
$z = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)$
$z = 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180$
$z = 2.5x + 1.5y + 410$
The problem is to minimize $z = 2.5x + 1.5y + 410$ subject to:
$x+y \leq 100, x \leq 60, y \leq 50, x+y \geq 60, x, y \geq 0$.
The corner points of the feasible region are $A(60, 0), B(60, 40), C(50, 50),$ and $D(10, 50)$.

Corner point	$z = 2.5x + 1.5y + 410$
$A(60, 0)$	$560$
$B(60, 40)$	$620$
$C(50, 50)$	$610$
$D(10, 50)$	$510$ (Minimum)

The minimum value of $z$ is $510$ at $(10, 50)$.
Thus,the amount of grain transported from $A$ to $D, E, F$ is $10, 50, 40$ quintals respectively,and from $B$ to $D, E, F$ is $50, 0, 0$ quintals respectively.

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.
The objective function is to maximize nitrogen: $Z = 3x + 3.5y$.
Constraints based on the table:
$1$. Phosphoric acid: $x + 2y \geq 240$
$2$. Potash: $3x + 1.5y \geq 270 \implies 2x + y \geq 180$
$3$. Chlorine: $1.5x + 2y \leq 310$
$4$. Non-negativity: $x, y \geq 0$
Solving the system of inequalities,the feasible region is bounded by the corner points $A(140, 50)$,$B(20, 140)$,and $C(40, 100)$.
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 maximum nitrogen is $595 \, kg$ at $x = 140$ and $y = 50$.

(A) Let $x$ and $y$ be the number of dolls of type $A$ and $B$ respectively that are produced per week.
The given problem can be formulated as follows:
Maximize $z = 12x + 16y$ .....$(1)$
Subject to the constraints:
$x + y \leq 1200$ .....$(2)$
$y \leq \frac{x}{2} \Rightarrow x \geq 2y$ .....$(3)$
$x - 3y \leq 600$ .....$(4)$
$x, y \geq 0$ .....$(5)$
The feasible region is determined by the system of constraints. The corner points of the feasible region are $O(0,0)$,$A(600,0)$,$B(1050,150)$,and $C(800,400)$.
The values of $z$ at these corner points are as follows:

Corner point	$z = 12x + 16y$
$O(0,0)$	$0$
$A(600,0)$	$7200$
$B(1050,150)$	$12(1050) + 16(150) = 12600 + 2400 = 15000$
$C(800,400)$	$12(800) + 16(400) = 9600 + 6400 = 16000$

The maximum value of $z$ is $16000$ at $(800, 400)$.
Thus,$800$ dolls of type $A$ and $400$ dolls of type $B$ should be produced weekly to maximize the profit.

Let the manufacturer produce $x$ units of type $A$ circuits and $y$ units of type $B$ circuits.
From the given information,we have the following corresponding constraint table:

Component	Type $A$ $(x)$	Type $B$ $(y)$	Maximum Stock
Resistors	$20$	$10$	$200$
Transistors	$10$	$20$	$120$
Capacitors	$10$	$30$	$150$
Profit	$Rs. 50$	$Rs. 60$	-

Thus,the objective function for profit is $Z = 50x + 60y$.
Now,we have the following mathematical model for the given problem:
Maximize $Z = 50x + 60y$
Subject to the constraints:
$20x + 10y \leq 200$ (Resistors constraint) $\Rightarrow 2x + y \leq 20$
$10x + 20y \leq 120$ (Transistors constraint) $\Rightarrow x + 2y \leq 12$
$10x + 30y \leq 150$ (Capacitors constraint) $\Rightarrow x + 3y \leq 15$
$x \geq 0, y \geq 0$ (Non-negativity constraints)
So,the $LPP$ is to maximize $Z = 50x + 60y$ subject to $2x + y \leq 20, x + 2y \leq 12, x + 3y \leq 15, x \geq 0, y \geq 0$.

(A) Let $x$ be the number of large vans and $y$ be the number of small vans.
The objective is to minimize the total cost $Z$. The cost of one large van is $Rs. 400$ and one small van is $Rs. 200$. Thus,the objective function is $Z = 400x + 200y$.
Constraints:
$1$. Total packages to be transported is at least $1200$: $200x + 80y \geq 1200$,which simplifies to $5x + 2y \geq 30$.
$2$. Total cost cannot exceed $Rs. 3000$: $400x + 200y \leq 3000$,which simplifies to $2x + y \leq 15$.
$3$. Number of large vans cannot exceed the number of small vans: $x \leq y$.
$4$. Non-negativity constraints: $x \geq 0, y \geq 0$.
Thus,the $LPP$ is:
Minimize $Z = 400x + 200y$
Subject to:
$5x + 2y \geq 30$
$2x + y \leq 15$
$x \leq y$
$x, y \geq 0$

(A) Let the company manufacture $x$ number of type $A$ sweaters and $y$ number of type $B$ sweaters.
The company spends at most $Rs. 72000$ a day. Therefore,$360x + 120y \leq 72000$.
Dividing by $120$,we get $3x + y \leq 600 \dots (i)$.
The company can make at most $300$ sweaters in total. Therefore,$x + y \leq 300 \dots (ii)$.
The number of sweaters of type $B$ cannot exceed the number of sweaters of type $A$ by more than $100$. This means $y - x \leq 100$,which can be written as $-x + y \leq 100 \dots (iii)$.
The company makes a profit of $Rs. 200$ for each type $A$ sweater and $Rs. 120$ for each type $B$ sweater. The objective function to maximize profit is $Z = 200x + 120y$.
Thus,the $LPP$ formulation is:
Maximize $Z = 200x + 120y$
Subject to constraints:
$3x + y \leq 600$
$x + y \leq 300$
$-x + y \leq 100$
$x \geq 0, y \geq 0$

(A) Let the man ride his motorcycle for a distance $x \, km$ at the speed of $50 \, km/h$ and for a distance $y \, km$ at the speed of $80 \, km/h$.
The cost of petrol for distance $x$ is $2x$ and for distance $y$ is $3y$. Since he has at most $Rs. \, 120$ to spend,the cost constraint is $2x + 3y \leq 120$.
The time taken to travel distance $x$ is $\frac{x}{50}$ hours and for distance $y$ is $\frac{y}{80}$ hours. Since he has at most $1$ hour,the time constraint is $\frac{x}{50} + \frac{y}{80} \leq 1$,which simplifies to $8x + 5y \leq 400$.
Since distance cannot be negative,$x \geq 0$ and $y \geq 0$.
The objective is to maximize the total distance $Z = x + y$.
Thus,the linear programming problem is:
Maximize $Z = x + y$
Subject to:
$2x + 3y \leq 120$
$8x + 5y \leq 400$
$x, y \geq 0$

(A) Let the manufacturer produce $x$ number of Model $X$ bikes and $y$ number of Model $Y$ bikes.
Model $X$ takes $6$ man-hours and Model $Y$ takes $10$ man-hours per unit. Total man-hours available per week is $450$.
$\therefore 6x + 10y \leq 450 \Rightarrow 3x + 5y \leq 225$
Handling and marketing costs are $Rs. 2000$ and $Rs. 1000$ per unit respectively,with total funds of $Rs. 80,000$ per week.
$\therefore 2000x + 1000y \leq 80000 \Rightarrow 2x + y \leq 80$
Also,$x \geq 0, y \geq 0$.
We need to maximize the profit function $Z = 1000x + 500y$ subject to the constraints:
$3x + 5y \leq 225$
$2x + y \leq 80$
$x \geq 0, y \geq 0$
The feasible region is determined by the corner points $(0,0), (40,0), (25,30),$ and $(0,45)$.

Corner Points	Value of $Z = 1000x + 500y$
$(0,0)$	$0$
$(40,0)$	$1000(40) + 500(0) = 40000$
$(25,30)$	$1000(25) + 500(30) = 25000 + 15000 = 40000$
$(0,45)$	$1000(0) + 500(45) = 22500$

The maximum profit is $Rs. 40,000$. This occurs at any point on the line segment joining $(40,0)$ and $(25,30)$. Thus,the manufacturer can produce $40$ units of Model $X$ and $0$ units of Model $Y$,or $25$ units of Model $X$ and $30$ units of Model $Y$ to achieve the maximum profit.

(C) Let the person take $x$ units of tablet $X$ and $y$ units of tablet $Y$.
From the given tabulated information,we have the following constraints:
$6x + 2y \geq 18 \Rightarrow 3x + y \geq 9$
$3x + 3y \geq 21 \Rightarrow x + y \geq 7$
$2x + 4y \geq 16 \Rightarrow x + 2y \geq 8$
Also,$x \geq 0, y \geq 0$.
The objective is to minimize the cost $Z = 2x + y$.
Plotting the inequalities,the feasible region is unbounded with corner points $A(8, 0)$,$B(3, 4)$,$C(1, 6)$,and $D(0, 9)$.

Corner points	Corresponding value of $Z = 2x + y$
$(8, 0)$	$16$
$(3, 4)$	$10$
$(1, 6)$	$8$ (Minimum)
$(0, 9)$	$9$

Since the feasible region is unbounded,we check the inequality $2x + y < 8$. The open half-plane defined by $2x + y < 8$ has no common points with the feasible region. Therefore,the minimum value of $Z$ is $8$ at the point $(1, 6)$.
Thus,the person should take $1$ unit of tablet $X$ and $6$ units of tablet $Y$ to satisfy the requirements at the minimum cost of $Rs. 8$.

(B) Let factory $I$ operate for $x$ days and factory $II$ operate for $y$ days.
At factory $I, 50$ calculators of model $A$ and at factory $II, 40$ calculators of model $A$ are made every day. The company has orders for at least $6400$ calculators of model $A$.
$\therefore 50x + 40y \geq 6400 \Rightarrow 5x + 4y \geq 640$
At factory $I, 50$ calculators of model $B$ and at factory $II, 20$ calculators of model $B$ are made every day. The company has orders for at least $4000$ calculators of model $B$.
$\therefore 50x + 20y \geq 4000 \Rightarrow 5x + 2y \geq 400$
At factory $I, 30$ calculators of model $C$ and at factory $II, 40$ calculators of model $C$ are made every day. The company has orders for at least $4800$ calculators of model $C$.
$\therefore 30x + 40y \geq 4800 \Rightarrow 3x + 4y \geq 480$
Also,$x \geq 0, y \geq 0.$
We have to minimize the cost $Z = 12000x + 15000y$ subject to the constraints:
$5x + 4y \geq 640$
$5x + 2y \geq 400$
$3x + 4y \geq 480$
$x, y \geq 0$
The feasible region is unbounded with corner points $A(160, 0), B(80, 60), C(32, 120),$ and $D(0, 200).$

Corner points	Value of $Z = 12000x + 15000y$
$(160, 0)$	$1920000$
$(80, 60)$	$1860000$ (Minimum)
$(32, 120)$	$2184000$
$(0, 200)$	$3000000$

To verify the minimum,we plot $12000x + 15000y < 1860000$ or $4x + 5y < 620.$ As shown in the figure,there are no common points with the feasible region,so the minimum value is $1860000.$
Thus,factory $I$ should operate for $80$ days and factory $II$ should operate for $60$ days.

(C) The given constraints are:
$1$) $-x + y \leq 1$
$2$) $2x + y \leq 2$
$3$) $x \geq 0, y \geq 0$
To find the feasible region,we plot the lines:
- For $-x + y = 1$,the intercepts are $(0, 1)$ and $(-1, 0)$.
- For $2x + y = 2$,the intercepts are $(0, 2)$ and $(1, 0)$.
The intersection point of $-x + y = 1$ and $2x + y = 2$ is found by subtracting the equations:
$(2x + y) - (-x + y) = 2 - 1 \implies 3x = 1 \implies x = 1/3$.
Substituting $x = 1/3$ into $2x + y = 2$:
$2(1/3) + y = 2 \implies y = 2 - 2/3 = 4/3$.
So,the intersection point is $(1/3, 4/3)$.
The corner points of the feasible region are $(0, 0)$,$(1, 0)$,and $(1/3, 4/3)$.
Now,evaluate $z = 2x + 6y$ at these points:
- At $(0, 0)$: $z = 2(0) + 6(0) = 0$.
- At $(1, 0)$: $z = 2(1) + 6(0) = 2$.
- At $(1/3, 4/3)$: $z = 2(1/3) + 6(4/3) = 2/3 + 24/3 = 26/3$.
The maximum value is $26/3$.

(A) The feasible region is determined by the constraints:
$1) 0 \leq x \leq 4$
$2) 1 \leq y \leq 6$
$3) x + y \leq 5$
To find the vertices of the feasible region,we solve the intersection points of the boundary lines:
- Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.
- Intersection of $x = 0$ and $x + y = 5$ is $(0, 5)$.
- Intersection of $y = 1$ and $x + y = 5$ is $(4, 1)$.
- Intersection of $x = 4$ and $y = 1$ is $(4, 1)$ (same as above).
- Intersection of $x = 4$ and $x + y = 5$ is $(4, 1)$.
The vertices of the feasible region are $(0, 1), (0, 5),$ and $(4, 1)$.
Now,evaluate $z = -3x + 2y$ at each vertex:
- At $(0, 1): z = -3(0) + 2(1) = 2$
- At $(0, 5): z = -3(0) + 2(5) = 10$
- At $(4, 1): z = -3(4) + 2(1) = -12 + 2 = -10$
The minimum value of $z$ is $-10$ at the point $(4, 1)$.

(A) Let the number of cake $A$ be $x$ and the number of cake $B$ be $y$.

Ingredient	Cake $A$ $(g)$	Cake $B$ $(g)$	Total Available $(g)$
Flour	$200$	$100$	$5000$
Fat	$25$	$50$	$1000$

Constraints for flour: $200x + 100y \leq 5000$. Dividing by $100$,we get $2x + y \leq 50$.
Constraints for fat: $25x + 50y \leq 1000$. Dividing by $25$,we get $x + 2y \leq 40$.
Non-negativity constraints: $x \geq 0, y \geq 0$.
Objective function to maximize total cakes: $Z = x + y$.
Thus,the mathematical form is $Z = x + y, 2x + y \leq 50, x + 2y \leq 40, x \geq 0, y \geq 0$.

(B) To find the maximum value of $Z = 3x_{1} + 5x_{2}$,we evaluate the objective function at the corner points of the feasible region defined by the constraints:
$1$. $3x_{1} + 2x_{2} \leq 18$
$2$. $x_{1} \leq 4$
$3$. $x_{2} \leq 6$
$4$. $x_{1} \geq 0, x_{2} \geq 0$
The corner points are found by the intersection of these lines:
- Intersection of $x_{1} = 0$ and $x_{2} = 0$ is $(0, 0)$. $Z = 0$.
- Intersection of $x_{1} = 4$ and $x_{2} = 0$ is $(4, 0)$. $Z = 3(4) + 5(0) = 12$.
- Intersection of $x_{1} = 4$ and $3x_{1} + 2x_{2} = 18$ gives $3(4) + 2x_{2} = 18 \Rightarrow 2x_{2} = 6 \Rightarrow x_{2} = 3$. Point is $(4, 3)$. $Z = 3(4) + 5(3) = 12 + 15 = 27$.
- Intersection of $3x_{1} + 2x_{2} = 18$ and $x_{2} = 6$ gives $3x_{1} + 2(6) = 18 \Rightarrow 3x_{1} = 6 \Rightarrow x_{1} = 2$. Point is $(2, 6)$. $Z = 3(2) + 5(6) = 6 + 30 = 36$.
- Intersection of $x_{1} = 0$ and $x_{2} = 6$ is $(0, 6)$. $Z = 3(0) + 5(6) = 30$.
Comparing the values of $Z$ at all corner points,the maximum value is $36$ at $(2, 6)$.

(D) To determine which point is not in the solution set,we test each point against the given constraints: $x+2y \leq 2000$,$x+y \leq 1500$,$y \leq 600$,and $x \geq 0$.
For $(1000, 0)$:
$1000 + 2(0) = 1000 \leq 2000$ (True)
$1000 + 0 = 1000 \leq 1500$ (True)
$0 \leq 600$ (True)
$1000 \geq 0$ (True)
This point is in the region.
For $(0, 500)$:
$0 + 2(500) = 1000 \leq 2000$ (True)
$0 + 500 = 500 \leq 1500$ (True)
$500 \leq 600$ (True)
$0 \geq 0$ (True)
This point is in the region.
For $(2, 0)$:
$2 + 2(0) = 2 \leq 2000$ (True)
$2 + 0 = 2 \leq 1500$ (True)
$0 \leq 600$ (True)
$2 \geq 0$ (True)
This point is in the region.
For $(2000, 0)$:
$2000 + 2(0) = 2000 \leq 2000$ (True)
$2000 + 0 = 2000 \leq 1500$ (False)
Since the condition $x+y \leq 1500$ is violated,the point $(2000, 0)$ is not in the solution set.

(C) The constraints are given by:
$1$) $2x + 3y \leq 6$
$2$) $x + 4y \leq 4$
$3$) $x \geq 0, y \geq 0$
To find the corner points of the feasible region,we identify the intersection points of the boundary lines:
- The line $2x + 3y = 6$ intersects the axes at $(3, 0)$ and $(0, 2)$.
- The line $x + 4y = 4$ intersects the axes at $(4, 0)$ and $(0, 1)$.
Solving the system of equations $2x + 3y = 6$ and $x + 4y = 4$:
From the second equation,$x = 4 - 4y$.
Substitute into the first: $2(4 - 4y) + 3y = 6 \implies 8 - 8y + 3y = 6 \implies -5y = -2 \implies y = \frac{2}{5}$.
Then $x = 4 - 4(\frac{2}{5}) = 4 - \frac{8}{5} = \frac{12}{5}$.
The corner points of the feasible region are $(0, 0)$,$(3, 0)$,$(0, 1)$,and $(\frac{12}{5}, \frac{2}{5})$.
Comparing these with the given options,the point $(\frac{12}{5}, \frac{2}{5})$ is a corner point.

(B) To determine the objective function for maximum profit,we consider the profit earned per unit of each commodity.
Let $x$ be the quantity of rice in quintals and $y$ be the quantity of wheat in quintals.
Profit from $x$ quintals of rice = $200 \times x = 200x$.
Profit from $y$ quintals of wheat = $125 \times y = 125y$.
The total profit $Z$ is the sum of the profits from rice and wheat.
Therefore,the objective function is $Z = 200x + 125y$.

(D) To find the maximum profit,we evaluate the objective function $z = 2000x + 5000y$ at each corner point of the feasible region:

Corner Point $(x, y)$	Value of $z = 2000x + 5000y$
$(1, 0)$	$z = 2000(1) + 5000(0) = 2000$
$(2, 0)$	$z = 2000(2) + 5000(0) = 4000$
$(0, 2)$	$z = 2000(0) + 5000(2) = 10000$ (Maximum)
$(0, 1)$	$z = 2000(0) + 5000(1) = 5000$

Comparing the values,the maximum profit is $10000$.

(A) To find the minimum value of $z = 2x + 4y$,we first identify the feasible region defined by the constraints:
$1$) $x + 2y \geq 10$
$2$) $3x + y \geq 10$
$3$) $x \geq 0, y \geq 0$
Find the intersection points of the boundary lines:
- For $x + 2y = 10$ and $3x + y = 10$:
Multiply the second equation by $2$: $6x + 2y = 20$.
Subtract the first equation: $(6x + 2y) - (x + 2y) = 20 - 10 \implies 5x = 10 \implies x = 2$.
Substitute $x = 2$ into $x + 2y = 10$: $2 + 2y = 10 \implies 2y = 8 \implies y = 4$.
Intersection point: $(2, 4)$.
- Intercepts for $x + 2y = 10$: $(10, 0)$ and $(0, 5)$.
- Intercepts for $3x + y = 10$: $(10/3, 0)$ and $(0, 10)$.
The corner points of the unbounded feasible region are $(0, 10)$,$(2, 4)$,and $(10, 0)$.
Evaluate $z = 2x + 4y$ at these corner points:
- At $(0, 10)$: $z = 2(0) + 4(10) = 40$.
- At $(2, 4)$: $z = 2(2) + 4(4) = 4 + 16 = 20$.
- At $(10, 0)$: $z = 2(10) + 4(0) = 20$.
Since the feasible region is unbounded,we check if $z < 20$ is possible. The line $2x + 4y = 20$ (or $x + 2y = 10$) is a boundary line. The minimum value is $20$.

(D) To find the minimum value of the objective function $F = 4x + 6y$,we evaluate $F$ at each corner point of the feasible region:

Corner Point $(x, y)$	Objective Function $F = 4x + 6y$
$(0, 2)$	$F = 4(0) + 6(2) = 12$
$(3, 0)$	$F = 4(3) + 6(0) = 12$
$(6, 0)$	$F = 4(6) + 6(0) = 24$
$(6, 8)$	$F = 4(6) + 6(8) = 72$
$(0, 5)$	$F = 4(0) + 6(5) = 30$

The minimum value of $F$ is $12$,which occurs at both corner points $(0, 2)$ and $(3, 0)$.
According to the property of linear programming,if the objective function has the same minimum value at two corner points,then it has the same minimum value at every point on the line segment joining these two points.
Therefore,the minimum value of $F$ occurs at any point on the line segment joining $(0, 2)$ and $(3, 0)$.

(A) The objective function is $z = 6xy + y^2$. The constraints are $3x + 4y \leq 100$,$4x + 3y \leq 75$,$x \geq 0$,and $y \geq 0$.
From the constraints,the feasible region is a polygon with vertices $(0, 0)$,$(18.75, 0)$,and $(0, 25)$.
Evaluating $z$ at the vertices:
At $(0, 0)$,$z = 6(0)(0) + 0^2 = 0$.
At $(18.75, 0)$,$z = 6(18.75)(0) + 0^2 = 0$.
At $(0, 25)$,$z = 6(0)(25) + 25^2 = 625$.
However,since $z$ is not a linear function,we check the boundary $4x + 3y = 75$,which implies $x = \frac{75-3y}{4}$.
Substituting this into $z$: $z = 6y(\frac{75-3y}{4}) + y^2 = \frac{3y(75-3y)}{2} + y^2 = \frac{225y - 9y^2 + 2y^2}{2} = \frac{225y - 7y^2}{2}$.
To find the maximum,take the derivative with respect to $y$ and set to $0$: $\frac{dz}{dy} = \frac{225 - 14y}{2} = 0 \implies y = \frac{225}{14} \approx 16.07$.
Then $x = \frac{75 - 3(225/14)}{4} = \frac{1050 - 675}{56} = \frac{375}{56} \approx 6.7$.
Substituting these values into $z$: $z = \frac{225(225/14) - 7(225/14)^2}{2} = \frac{50625/14 - 354375/196}{2} = \frac{708750 - 354375}{392} = \frac{354375}{392} \approx 904.0178$.
The maximum value is approximately $904$.

(A) We are given the objective function $Z = 5x + 4y$ subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,and $y \geq 0$.
First,we identify the feasible region by plotting the lines $y = 2x$,$x = 2y$,and $x+y = 3$.
$1$. The intersection of $y = 2x$ and $x+y = 3$: Substituting $y=2x$ into $x+y=3$ gives $x+2x=3$,so $3x=3$,which means $x=1$. Then $y=2(1)=2$. So,the point is $A(1, 2)$.
$2$. The intersection of $x = 2y$ and $x+y = 3$: Substituting $x=2y$ into $x+y=3$ gives $2y+y=3$,so $3y=3$,which means $y=1$. Then $x=2(1)=2$. So,the point is $B(2, 1)$.
$3$. The origin $O(0, 0)$ is also a corner point.
The feasible region is the triangle $OAB$. We evaluate $Z$ at the corner points:

Corner Point	$Z = 5x + 4y$
$O(0, 0)$	$5(0) + 4(0) = 0$
$A(1, 2)$	$5(1) + 4(2) = 5 + 8 = 13$
$B(2, 1)$	$5(2) + 4(1) = 10 + 4 = 14$

Thus,the maximum value of $Z$ is $14$ at point $B(2, 1)$.

(C) The objective function is $z = ax + by$ with $a, b > 0$. The constraints are $x \leq 2, y \leq 2, x + y \geq 3, x \geq 0, y \geq 0$.
Plotting these constraints,we identify the feasible region as the triangle with vertices $A(2, 1)$,$B(1, 2)$,and $C(2, 2)$.
Since the minimum value of $z$ occurs only at $(2, 1)$,the value of $z$ at $(2, 1)$ must be strictly less than the value of $z$ at the other corner points.
Comparing $z$ at $A(2, 1)$ and $B(1, 2)$:
$z(2, 1) = 2a + b$
$z(1, 2) = a + 2b$
For the minimum to be at $(2, 1)$,we must have $z(2, 1) < z(1, 2)$.
$2a + b < a + 2b$
$2a - a < 2b - b$
$a < b$
Thus,the correct condition is $a < b$.

(B) We are given the objective function: Maximize $z = 4x_1 + 2x_2$.
Subject to the constraints:
$1) 3x_1 + 2x_2 \geq 9$
$2) x_1 - x_2 \leq 3$
$3) x_1 \geq 0, x_2 \geq 0$
By plotting these lines on the coordinate plane:
For $3x_1 + 2x_2 = 9$,the intercepts are $(3, 0)$ and $(0, 4.5)$. The region is away from the origin.
For $x_1 - x_2 = 3$,the intercepts are $(3, 0)$ and $(0, -3)$. The region is towards the origin.
Upon analyzing the feasible region,we observe that the region is unbounded in the first quadrant. For an unbounded feasible region,if the objective function increases indefinitely as we move along the boundary,the $L$.$P$.$P$. has an unbounded solution. Here,as $x_1$ and $x_2$ increase,$z = 4x_1 + 2x_2$ can take arbitrarily large values within the feasible region. Therefore,the $L$.$P$.$P$. has an unbounded solution.

(A) To find the minimum value of the objective function $z = 10x + 25y$,we first identify the feasible region defined by the constraints $0 \leq x \leq 3$,$0 \leq y \leq 3$,and $x + y \geq 5$.
The vertices of the feasible region are determined by the intersection of the lines:
$1$. Intersection of $x = 3$ and $x + y = 5$ gives $y = 2$,so the point is $(3, 2)$.
$2$. Intersection of $x = 3$ and $y = 3$ gives the point $(3, 3)$.
$3$. Intersection of $y = 3$ and $x + y = 5$ gives $x = 2$,so the point is $(2, 3)$.
Now,we evaluate $z = 10x + 25y$ at these corner points:
- At $(3, 2)$: $z = 10(3) + 25(2) = 30 + 50 = 80$.
- At $(3, 3)$: $z = 10(3) + 25(3) = 30 + 75 = 105$.
- At $(2, 3)$: $z = 10(2) + 25(3) = 20 + 75 = 95$.
Comparing these values,the minimum value is $80$.

(D) The given constraints are $3x+2y \leq 12$,$2x+3y \leq 12$,$x \geq 0$,and $y \geq 0$.
To find the feasible region,we identify the corner points of the region bounded by these inequalities.
The lines are $L_1: 3x+2y=12$ and $L_2: 2x+3y=12$.
The intersection point $E$ of $L_1$ and $L_2$ is found by solving the system:
$3x+2y=12$ (multiply by $3$) $\Rightarrow 9x+6y=36$
$2x+3y=12$ (multiply by $2$) $\Rightarrow 4x+6y=24$
Subtracting gives $5x=12$,so $x=2.4$.
Substituting $x=2.4$ into $3(2.4)+2y=12$ gives $7.2+2y=12$,so $2y=4.8$,$y=2.4$.
The corner points of the feasible region are $(0,0)$,$(4,0)$,$(0,4)$,and $(2.4, 2.4)$.
Now evaluate $z=9x+11y$ at these points:
$z(0,0) = 9(0)+11(0) = 0$
$z(4,0) = 9(4)+11(0) = 36$
$z(0,4) = 9(0)+11(4) = 44$
$z(2.4, 2.4) = 9(2.4)+11(2.4) = 21.6+26.4 = 48$
The maximum value is $48$.

(A) To find the maximum value of $Z = 2x + y$,we identify the corner points of the feasible region defined by the constraints $3x + 5y \leq 26$,$5x + 3y \leq 30$,$x \geq 0$,and $y \geq 0$.
$1$. Find the intersection of the lines $3x + 5y = 26$ and $5x + 3y = 30$:
Multiply the first equation by $5$ and the second by $3$:
$15x + 25y = 130$
$15x + 9y = 90$
Subtracting the equations: $16y = 40 \implies y = \frac{40}{16} = 2.5$.
Substitute $y = 2.5$ into $3x + 5(2.5) = 26 \implies 3x + 12.5 = 26 \implies 3x = 13.5 \implies x = 4.5$.
So,point $B$ is $(4.5, 2.5)$.
$2$. The corner points of the feasible region are $(0, 0)$,$(6, 0)$,$(4.5, 2.5)$,and $(0, 5.2)$.
$3$. Evaluate $Z = 2x + y$ at each corner point:
At $(0, 0): Z = 2(0) + 0 = 0$
At $(6, 0): Z = 2(6) + 0 = 12$
At $(4.5, 2.5): Z = 2(4.5) + 2.5 = 9 + 2.5 = 11.5$
At $(0, 5.2): Z = 2(0) + 5.2 = 5.2$
The maximum value is $12$ at the point $(6, 0)$.

(A) The objective function is $Z = 4 x_1 + 5 x_2$.
The constraints are:
$1) 2 x_1 + x_2 \geq 7$
$2) 2 x_1 + 3 x_2 \leq 15$
$3) x_2 \leq 3$
$4) x_1, x_2 \geq 0$
To find the feasible region,we determine the intersection points of the boundary lines:
For $2 x_1 + x_2 = 7$,the intercepts are $(3.5, 0)$ and $(0, 7)$.
For $2 x_1 + 3 x_2 = 15$,the intercepts are $(7.5, 0)$ and $(0, 5)$.
The line $x_2 = 3$ is a horizontal line.
Solving for intersection points of the feasible region:
- Intersection of $2 x_1 + x_2 = 7$ and $x_2 = 3$: $2 x_1 + 3 = 7 \implies 2 x_1 = 4 \implies x_1 = 2$. Point: $(2, 3)$.
- Intersection of $2 x_1 + 3 x_2 = 15$ and $x_2 = 3$: $2 x_1 + 9 = 15 \implies 2 x_1 = 6 \implies x_1 = 3$. Point: $(3, 3)$.
- The $x_1$-intercepts are $(3.5, 0)$ and $(7.5, 0)$.
The corner points of the feasible region are $(3.5, 0), (7.5, 0), (3, 3), (2, 3)$.
Evaluating $Z = 4 x_1 + 5 x_2$ at these points:

Corner Point	$Z = 4 x_1 + 5 x_2$
$(3.5, 0)$	$4(3.5) + 5(0) = 14$
$(7.5, 0)$	$4(7.5) + 5(0) = 30$
$(3, 3)$	$4(3) + 5(3) = 12 + 15 = 27$
$(2, 3)$	$4(2) + 5(3) = 8 + 15 = 23$

The minimum value of $Z$ is $14$,which occurs at the point $(3.5, 0)$. Since the $x_2$-coordinate is $0$,this point lies on the $x_1$-axis.