Word problem of Linear programming Questions in English

(C) The feasible region is determined by the constraints $x_1+x_2 \leq 10$,$-2x_1+3x_2 \leq 15$,$x_1 \leq 6$,and $x_1, x_2 \geq 0$. The corner points of the feasible region are $O(0,0)$,$E(6,0)$,$F(6,4)$,$G(3,7)$,and $D(0,5)$.
We evaluate the objective function $z=x_1+x_2$ at these corner points:
$z(O) = 0+0 = 0$
$z(E) = 6+0 = 6$
$z(F) = 6+4 = 10$
$z(G) = 3+7 = 10$
$z(D) = 0+5 = 5$
The maximum value of $z$ is $10$,which occurs at both corner points $F(6,4)$ and $G(3,7)$.
Since the objective function is maximized at two distinct corner points,it will have the same maximum value at every point on the line segment joining these two points.

(D) To find the feasible region,we analyze the given constraints:
$1$. $2x + 3y \geqslant 12$: The region is on or above the line $2x + 3y = 12$.
$2$. $-x + y \leqslant 3$: The region is on or below the line $-x + y = 3$.
$3$. $x \leqslant 4$: The region is to the left of the line $x = 4$.
$4$. $y \geqslant 3$: The region is on or above the line $y = 3$.
By observing the graph and testing the constraints:
- The region $S_1$ is above $y=3$,to the left of $x=4$,above $2x+3y=12$,and above $-x+y=3$. This does not satisfy the constraints.
- The region $S_4$ is bounded by $y=3$,$x=4$,$-x+y=3$,and $2x+3y=12$. Checking a point in $S_4$,say $(1, 4)$:
- $2(1) + 3(4) = 14 \geqslant 12$ (True)
- $-(1) + 4 = 3 \leqslant 3$ (True)
- $1 \leqslant 4$ (True)
- $4 \geqslant 3$ (True)
All constraints are satisfied in region $S_4$.

(B) The given constraints are:
$1$) $x - 2 \leqslant y \implies y \geqslant x - 2$
$2$) $x \geqslant y - 1 \implies y \leqslant x + 1$
$3$) $x \geqslant 2$
$4$) $y \leqslant 4$
$5$) $x, y \geqslant 0$
Analyzing these inequalities:
- The line $y = x - 2$ passes through $(2, 0)$ and $(4, 2)$. The region $y \geqslant x - 2$ is above this line.
- The line $y = x + 1$ passes through $(0, 1)$ and $(3, 4)$. The region $y \leqslant x + 1$ is below this line.
- The constraint $x \geqslant 2$ restricts the region to the right of the vertical line $x = 2$.
- The constraint $y \leqslant 4$ restricts the region below the horizontal line $y = 4$.
Combining these,the feasible region is bounded by $x = 2$,$y = x - 2$,$y = x + 1$,and $y = 4$. This corresponds to the shaded region in option $B$.

(A) The feasible region is bounded by the vertices $A, B, C$.
From the graph,the lines are $x = 5$,$y = 4$,$x + y = 15$,and $x - y = 10$.
$1$. Vertex $A$ is the intersection of $x = 5$ and $x - y = 10$. Substituting $x = 5$,we get $5 - y = 10$,so $y = -5$. However,the region is in the first quadrant,so we look at the intersection of $x = 5$ and $x + y = 15$,which is $(5, 10)$.
$2$. Vertex $B$ is the intersection of $x = 5$ and $x + y = 15$,which is $(5, 10)$.
$3$. Vertex $C$ is the intersection of $y = 4$ and $x + y = 15$,which is $(11, 4)$.
$4$. Vertex $D$ is the intersection of $y = 4$ and $x - y = 10$,which is $(14, 4)$.
Wait,looking at the shaded region,the vertices are $A(5, 5)$,$B(5, 10)$,and $C(11, 4)$.
- At $A(5, 5)$: $z = 550(5) + 300(5) = 2750 + 1500 = 4250$.
- At $B(5, 10)$: $z = 550(5) + 300(10) = 2750 + 3000 = 5750$.
- At $C(11, 4)$: $z = 550(11) + 300(4) = 6050 + 1200 = 7250$.
The maximum value is $7250$.

(B) To solve the $L$.$P$.$P$.,we identify the feasible region defined by the constraints:
$1$. $x + y \leqslant 8$
$2$. $x + 2y \geqslant 4$
$3$. $6x + 4y \geqslant 12$ (or $3x + 2y \geqslant 6$)
$4$. $x \geqslant 0, y \geqslant 0$
The corner points of the feasible region are determined by the intersection of these lines:
- Intersection of $x + 2y = 4$ and $3x + 2y = 6$: Subtracting gives $2x = 2$,so $x = 1$. Then $1 + 2y = 4 \implies y = 1.5$. Point: $(1, 1.5)$.
- Intersection of $x + y = 8$ and $x + 2y = 4$: $y = -4$,which is outside the first quadrant.
- Intersection of $x + y = 8$ and $3x + 2y = 6$: $y = -18$,outside the first quadrant.
- Boundary points on axes: $(0, 3)$ from $3x + 2y = 6$,$(0, 2)$ from $x + 2y = 4$,$(8, 0)$ from $x + y = 8$,$(2, 0)$ from $3x + 2y = 6$.
Evaluating $z = 30x + 20y$ at corner points:
- At $(0, 3)$,$z = 30(0) + 20(3) = 60$.
- At $(1, 1.5)$,$z = 30(1) + 20(1.5) = 30 + 30 = 60$.
- At $(8, 0)$,$z = 30(8) + 20(0) = 240$.
- At $(2, 0)$,$z = 30(2) + 20(0) = 60$.
Since the objective function $z$ takes the same minimum value of $60$ at multiple points on the line segment connecting $(0, 3)$ and $(2, 0)$ (specifically,the segment between $(0, 3)$ and $(1, 1.5)$ and $(1, 1.5)$ and $(2, 0)$),the $L$.$P$.$P$. has infinitely many solutions.

(C) To minimize $z = x + y$ subject to the constraints:
$1$) $x + y \geqslant 2$
$2$) $x + 2y \leqslant 8$
$3$) $y \leqslant 3$
$4$) $x, y \geqslant 0$
First,we identify the feasible region by plotting the lines:
- $x + y = 2$ passes through $(2, 0)$ and $(0, 2)$.
- $x + 2y = 8$ passes through $(8, 0)$ and $(0, 4)$.
- $y = 3$ is a horizontal line.
The vertices of the feasible region are found by the intersection of these lines:
- Intersection of $x + y = 2$ and $x = 0$ is $(0, 2)$.
- Intersection of $x + y = 2$ and $y = 0$ is $(2, 0)$.
- Intersection of $x + 2y = 8$ and $y = 0$ is $(8, 0)$ (but this is outside the region $y \leqslant 3$ and $x+y \geqslant 2$ constraints).
- Intersection of $x + 2y = 8$ and $y = 3$ is $x + 6 = 8 \implies x = 2$,so $(2, 3)$.
- Intersection of $x = 0$ and $y = 3$ is $(0, 3)$.
The corner points of the feasible region are $(0, 2), (2, 0), (2, 3), (0, 3)$.
Evaluating $z = x + y$ at these points:
- At $(0, 2)$,$z = 0 + 2 = 2$.
- At $(2, 0)$,$z = 2 + 0 = 2$.
- At $(2, 3)$,$z = 2 + 3 = 5$.
- At $(0, 3)$,$z = 0 + 3 = 3$.
The minimum value is $2$,which occurs at all points on the line segment connecting $(0, 2)$ and $(2, 0)$. Since a line segment contains infinitely many points,the solution set contains infinitely many points.

(C) The constraints are $x + y \leqslant 20$,$y \geqslant 5$,and $x \geqslant 0$.
The feasible region is a triangle with vertices determined by the intersection of these lines:
$1$. Intersection of $y = 5$ and $x = 0$: $(0, 5)$.
$2$. Intersection of $x + y = 20$ and $y = 5$: $(15, 5)$.
$3$. Intersection of $x + y = 20$ and $x = 0$: $(0, 20)$.
Now,evaluate $z = 7x - 8y$ at these vertices:
At $(0, 5)$: $z = 7(0) - 8(5) = -40$.
At $(15, 5)$: $z = 7(15) - 8(5) = 105 - 40 = 65$.
At $(0, 20)$: $z = 7(0) - 8(20) = -160$.
The maximum value is $65$ and the minimum value is $-160$.
The difference is $65 - (-160) = 65 + 160 = 225$.
Given the difference is $5k + 200$,we have $5k + 200 = 225$.
$5k = 25$,so $k = 5$.

(C) To find the constraints,we identify the equations of the lines $L_1, L_2, L_3$ from the graph.
$1$. Line $L_1$ passes through $(-1, 0)$ and $(0, 1)$. The equation is $\frac{x}{-1} + \frac{y}{1} = 1 \Rightarrow y-x = 1$. Since the feasible region is above this line,the constraint is $y-x \geqslant 1$.
$2$. Line $L_2$ passes through $(0, 2)$ and $(5, 0)$. The equation is $\frac{x}{5} + \frac{y}{2} = 1 \Rightarrow 2x+5y = 10$. Since the region is below this line,the constraint is $2x+5y \leqslant 10$.
$3$. Line $L_3$ passes through $(0, 1)$ and $(1, 0)$. The equation is $\frac{x}{1} + \frac{y}{1} = 1 \Rightarrow x+y = 1$. Since the region is above this line,the constraint is $x+y \geqslant 1$.
Combining these with non-negativity constraints $x, y \geqslant 0$,we get the system: $y-x \geqslant 1, 2x+5y \leqslant 10, x+y \geqslant 1, x, y \geqslant 0$. This matches option $C$.

(C) To find the feasible region,we analyze the given constraints:
$1. 2x + y \leqslant 10$: The line passes through $(0, 10)$ and $(5, 0)$. The region is towards the origin.
$2. y \leqslant x$: The line passes through $(0, 0)$ and $(2, 2)$. The region is below the line.
$3. y \leqslant 2$: The region is below the horizontal line $y = 2$.
$4. x, y \geqslant 0$: The region is in the first quadrant.
Intersection points:
- For $y = 2$ and $y = x$,we get $x = 2$. Point is $(2, 2)$.
- For $y = 2$ and $2x + y = 10$,we get $2x + 2 = 10 \Rightarrow 2x = 8 \Rightarrow x = 4$. Point is $(4, 2)$.
- For $y = x$ and $2x + y = 10$,we get $2x + x = 10 \Rightarrow 3x = 10 \Rightarrow x = 10/3$. Point is $(10/3, 10/3)$.
The feasible region is bounded by the vertices $(0, 0)$,$(2, 2)$,$(4, 2)$,and the $x$-axis segment. Looking at the provided options,the graph that correctly represents the region bounded by these lines is shown in option $C$.

(D) The constraints are $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x = y$,and $x, y \geqslant 0$.
Substituting $x = y$ into the constraints:
$1$) $x + 3x \leqslant 60 \implies 4x \leqslant 60 \implies x \leqslant 15$.
$2$) $x + x \geqslant 10 \implies 2x \geqslant 10 \implies x \geqslant 5$.
Since $x = y$,the feasible region is the line segment from $(5, 5)$ to $(15, 15)$.
The objective function is $z = 3x + 5y$.
Substituting $y = x$ into $z$,we get $z = 3x + 5x = 8x$.
At $(5, 5)$,$z = 8(5) = 40$.
At $(15, 15)$,$z = 8(15) = 120$.
The maximum value is $120$ and the minimum value is $40$.
The difference is $120 - 40 = 80$.

(B) To find the maximum value of $Z = 6x + 3y$,we first identify the feasible region defined by the constraints:
$1) x + y = 5$
$2) x + 2y = 4$
$3) 4x + y = 12$
$4) x \geq 0, y \geq 0$
Finding the intersection points of the lines:
- Intersection of $x + y = 5$ and $4x + y = 12$: Subtracting gives $3x = 7 \implies x = 7/3$. Then $y = 5 - 7/3 = 8/3$. Point: $(7/3, 8/3)$.
- Intersection of $x + 2y = 4$ and $x + y = 5$: Subtracting gives $y = -1$,which is outside the first quadrant.
- Intersection of $x + 2y = 4$ and $4x + y = 12$: $x = 4 - 2y$. Substituting: $4(4 - 2y) + y = 12 \implies 16 - 8y + y = 12 \implies 7y = 4 \implies y = 4/7$. Then $x = 4 - 8/7 = 20/7$. Point: $(20/7, 4/7)$.
- Boundary points on axes: $(3, 0)$ from $4x+y=12$,$(4, 0)$ from $x+2y=4$,$(0, 2)$ from $x+2y=4$,$(0, 5)$ from $x+y=5$.
Evaluating $Z = 6x + 3y$ at vertices of the feasible region:
- At $(3, 0): Z = 6(3) + 3(0) = 18$
- At $(7/3, 8/3): Z = 6(7/3) + 3(8/3) = 14 + 8 = 22$
- At $(20/7, 4/7): Z = 6(20/7) + 3(4/7) = (120 + 12)/7 = 132/7 \approx 18.85$
- At $(0, 2): Z = 6(0) + 3(2) = 6$
The maximum value is $22$.

(B) Let $x$ be the number of items $A$ and $y$ be the number of items $B$.
The profit function to be maximized is $z = 25x + 18y$.
Machine $I$ constraint: $20x + 15y \leqslant 640$ (since $10$ hours $40$ minutes $= 640$ minutes).
Machine $II$ constraint: $5x + 8y \leqslant 500$ (since $8$ hours $20$ minutes $= 500$ minutes).
Non-negativity constraints: $x \geqslant 0, y \geqslant 0$.
Thus,the correct formulation is: Maximize $z = 25x + 18y$ subject to $20x + 15y \leqslant 640, 5x + 8y \leqslant 500, x, y \geqslant 0$.

(A) The feasible region is determined by the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y \leqslant 0$,$x \geqslant 0$,and $y \geqslant 0$.
First,we find the vertices of the feasible region by solving the intersection points of the boundary lines:
$1$. Intersection of $x - y = 0$ and $x + y = 10$: $2x = 10 \implies x = 5, y = 5$. Vertex: $(5, 5)$.
$2$. Intersection of $x - y = 0$ and $x + 3y = 60$: $4y = 60 \implies y = 15, x = 15$. Vertex: $(15, 15)$.
$3$. Intersection of $x + y = 10$ and $x = 0$: $y = 10$. Vertex: $(0, 10)$.
$4$. Intersection of $x + 3y = 60$ and $x = 0$: $3y = 60 \implies y = 20$. Vertex: $(0, 20)$.
Now,we evaluate $Z = 3x + 5y$ at each vertex:
- At $(5, 5)$: $Z = 3(5) + 5(5) = 15 + 25 = 40$.
- At $(15, 15)$: $Z = 3(15) + 5(15) = 45 + 75 = 120$.
- At $(0, 10)$: $Z = 3(0) + 5(10) = 50$.
- At $(0, 20)$: $Z = 3(0) + 5(20) = 100$.
The maximum value is $120$ and the minimum value is $40$.
The difference is $120 - 40 = 80$.

(B) To find the feasible region,we analyze the given constraints:
$1$. $x - y \geqslant 0 \implies y \leqslant x$. This represents the region on or below the line $y = x$.
$2$. $x - 5y \leqslant -5 \implies 5y \geqslant x + 5 \implies y \geqslant \frac{1}{5}x + 1$. This represents the region on or above the line $y = \frac{1}{5}x + 1$.
$3$. $x \geqslant 0$ and $y \geqslant 0$ restrict the region to the first quadrant.
Combining these,we look for the region that is below $y = x$,above $y = \frac{1}{5}x + 1$,and in the first quadrant.
Solving for the intersection point: $x = \frac{1}{5}x + 1 \implies \frac{4}{5}x = 1 \implies x = 1.25$. Then $y = 1.25$. The intersection is at $(1.25, 1.25)$.
The region bounded by these lines in the first quadrant is a triangular region. Looking at the provided options,the figure corresponding to the region bounded by these inequalities is represented in option $B$.

(B) The feasible region lies on the origin side of the lines $x + y = 12$ and $2x + y = 20$ and is in the first quadrant.
The corner points of the feasible region are $O(0, 0)$,$B(10, 0)$,$C(8, 4)$,and $D(0, 12)$.
We evaluate the objective function $z = 10x + 6y$ at each corner point:
At $O(0, 0)$,$z = 10(0) + 6(0) = 0$.
At $B(10, 0)$,$z = 10(10) + 6(0) = 100$.
At $C(8, 4)$,$z = 10(8) + 6(4) = 80 + 24 = 104$.
At $D(0, 12)$,$z = 10(0) + 6(12) = 72$.
Therefore,the maximum value of $z$ is $104$,which occurs at the point $C(8, 4)$.

(A) To determine the system of linear inequalities,we identify the boundary lines of the shaded region:
$1$. The line passing through $(0, 25)$ and $(50, 0)$ has the equation $\frac{x}{50} + \frac{y}{25} = 1$,which simplifies to $x + 2y = 50$. Since the shaded region is above this line,the inequality is $x + 2y \geq 50$.
$2$. The line passing through $(0, 100)$ and $(50, 0)$ has the equation $\frac{x}{50} + \frac{y}{100} = 1$,which simplifies to $2x + y = 100$. Since the shaded region is below this line,the inequality is $2x + y \leq 100$.
$3$. The line passing through $(0, 0)$ and $(10, 20)$ has the equation $y = 2x$,which simplifies to $2x - y = 0$. Since the shaded region is to the left of this line (e.g.,test point $(0, 10)$ gives $2(0) - 10 = -10 \leq 0$),the inequality is $2x - y \leq 0$.
$4$. The region is in the first quadrant,so $x, y \geq 0$.
Thus,the correct set of inequalities is $x + 2y \geq 50, 2x + y \leq 100, 2x - y \leq 0, x, y \geq 0$.

(D) To determine the linear constraints,we analyze the boundaries of the shaded region:
$1$. The vertical line passing through $x=1$ with the shaded region to the right gives the constraint $x \geqslant 1$.
$2$. The horizontal line passing through $y=3$ with the shaded region below it gives the constraint $y \leqslant 3$.
$3$. The line passing through $(2, 0)$ and $(0, -1)$ has the equation $\frac{x}{2} - \frac{y}{1} = 1$,which simplifies to $x - 2y = 2$. Since the shaded region is above this line (e.g.,testing point $(3, 1)$ gives $3 - 2(1) = 1 \leqslant 2$),the constraint is $x - 2y \leqslant 2$.
$4$. The line passing through $(7, 0)$ and $(0, 6)$ has the equation $\frac{x}{7} + \frac{y}{6} = 1$,which simplifies to $6x + 7y = 42$. Since the shaded region is below this line (e.g.,testing point $(1, 1)$ gives $6(1) + 7(1) = 13 \leqslant 42$),the constraint is $6x + 7y \leqslant 42$.
$5$. The non-negativity constraints are $x \geqslant 0$ and $y \geqslant 0$.
Combining these,the correct set of constraints is $x \geqslant 1, y \leqslant 3, x-2y \leqslant 2, 6x+7y \leqslant 42, x \geqslant 0, y \geqslant 0$.

(D) The constraints are $3x+2y \leq 12$,$x+y \geq 4$,and $x, y \geq 0$.
To find the feasible region,we plot the lines $3x+2y=12$ and $x+y=4$.
The intersection points of the lines with the axes are:
For $3x+2y=12$: $(4,0)$ and $(0,6)$.
For $x+y=4$: $(4,0)$ and $(0,4)$.
The feasible region is the triangle with vertices $A(4,0)$,$B(0,4)$,and $C(0,6)$.
We evaluate the objective function $z=4x+6y$ at these corner points:
At $A(4,0)$: $z = 4(4) + 6(0) = 16$.
At $B(0,4)$: $z = 4(0) + 6(4) = 24$.
At $C(0,6)$: $z = 4(0) + 6(6) = 36$.
Comparing these values,the maximum value of $z$ is $36$,which occurs at point $C(0,6)$.

(C) The feasible region is bounded by the lines $x+y=10$,$5x+3y=15$,$x=6$,and the axes $x=0, y=0$.
The corner points of the feasible region are $A(0,5)$,$B(0,10)$,$C(6,4)$,and $E(3,0)$.
We evaluate the objective function $z=x+y$ at these corner points:
At $A(0,5)$,$z = 0+5 = 5$.
At $B(0,10)$,$z = 0+10 = 10$.
At $C(6,4)$,$z = 6+4 = 10$.
At $E(3,0)$,$z = 3+0 = 3$.
The maximum value of $z$ is $10$,which occurs at both points $B(0,10)$ and $C(6,4)$.
Since the maximum value occurs at two corner points,it occurs at all points on the line segment joining $B$ and $C$.
Therefore,the maximum value occurs at infinitely many points.

(B) The feasible region is bounded by the lines $3x+4y=12$,$x+y=5$,and the axes in the first quadrant.
To find the corner points,we identify the vertices of the shaded region:
$1$. Intersection of $3x+4y=12$ and the $x$-axis $(y=0)$: $3x=12 \implies x=4$. Point is $A(4, 0)$.
$2$. Intersection of $x+y=5$ and the $x$-axis $(y=0)$: $x=5$. Point is $B(5, 0)$.
$3$. Intersection of $x+y=5$ and the $y$-axis $(x=0)$: $y=5$. Point is $C(0, 5)$.
$4$. Intersection of $3x+4y=12$ and the $y$-axis $(x=0)$: $4y=12 \implies y=3$. Point is $D(0, 3)$.
Now,we evaluate $z=4x+2y$ at each corner point:
- At $A(4, 0): z = 4(4) + 2(0) = 16$
- At $B(5, 0): z = 4(5) + 2(0) = 20$
- At $C(0, 5): z = 4(0) + 2(5) = 10$
- At $D(0, 3): z = 4(0) + 2(3) = 6$
The maximum value of $z$ is $20$ at point $B(5, 0)$.

(D) To find the linear constraints,we identify the equations of the three lines forming the boundary of the shaded region:
$1$. Line passing through $(0, 3)$ and $(8, 0)$: The intercept form is $\frac{x}{8} + \frac{y}{3} = 1$,which simplifies to $3x + 8y = 24$. Since the shaded region is above this line (away from the origin),the constraint is $3x + 8y \geq 24$.
$2$. Line passing through $(0, 4)$ and $(5, 0)$: The intercept form is $\frac{x}{5} + \frac{y}{4} = 1$,which simplifies to $4x + 5y = 20$. Since the shaded region is below this line (towards the origin),the constraint is $4x + 5y \leq 20$.
$3$. Line passing through $(0, 5)$ and $(3, 0)$: The intercept form is $\frac{x}{3} + \frac{y}{5} = 1$,which simplifies to $5x + 3y = 15$. Since the shaded region is below this line (towards the origin),the constraint is $5x + 3y \leq 15$.
Combining these with the non-negativity constraints $x \geq 0, y \geq 0$,we get the system: $3x + 8y \geq 24, 4x + 5y \leq 20, 5x + 3y \leq 15, x \geq 0, y \geq 0$. Thus,option $D$ is correct.

(B) To find the graphical solution set,we analyze the given system of inequations: $x+y \geq 1$,$7x+9y \leq 63$,$y \leq 5$,$x \leq 6$,$x \geq 0$,$y \geq 0$.
$1$. For $x+y \geq 1$: The line passes through $(1, 0)$ and $(0, 1)$. The region is away from the origin.
$2$. For $7x+9y \leq 63$: The line passes through $(9, 0)$ and $(0, 7)$. The region is towards the origin.
$3$. For $x \leq 6$ and $y \leq 5$: These represent the region bounded by the lines $x=6$ and $y=5$ in the first quadrant.
$4$. The intersection of all these regions gives the feasible region. By plotting these lines,we observe that the region is bounded by the vertices $(0, 1), (0, 7), (2.57, 5), (6, 5), (6, 2.33)$ and $(1, 0)$.
Comparing this with the provided figures,the correct graphical representation is Fig. $2$.

(C) Let's analyze the shaded region and its linear constraints:
$1$. Looking at the graph,the region is bounded by the lines $y=x$ and $-x+3y=3$,along with the axes $x=0$ and $y=0$.
$2$. The shaded region is situated below the line $-x+3y=3$. Testing the origin $(0,0)$,we get $-0+3(0) = 0 \leq 3$,which satisfies the inequality. Thus,the constraint is $-x+3y \leq 3$.
$3$. The region is situated to the right of the line $y=x$ (or below the line $y=x$ in terms of $y$-values for a fixed $x$). Testing a point in the region,such as $(2, 0)$,we have $x-y = 2-0 = 2 \geq 0$. Wait,let's re-evaluate: for the region to the right of $y=x$,we have $x \geq y$,which means $x-y \geq 0$.
$4$. Re-checking the graph: The shaded region is bounded by $y=x$ on the left and $-x+3y=3$ on the top. For points in the region,$x \geq y$ (so $x-y \geq 0$) and $-x+3y \leq 3$.
$5$. Comparing with the options,option $(C)$ matches these conditions: $x-y \geq 0, -x+3y \leq 3, x \geq 0, y \geq 0$.

(B) Let $x$ and $y$ denote the quantity of copper and brass in grams,respectively.
The objective function to minimize is the cost: $z = 8x + 5y$.
The constraints based on the problem are:
$1) \ x + y \geq 5$ (Total weight at least $5 \text{ gms}$)
$2) \ x \geq 2$ (Minimum $2 \text{ gms}$ of copper)
$3) \ y \leq 4$ (Maximum $4 \text{ gms}$ of brass)
$4) \ x \geq 0, y \geq 0$ (Non-negativity constraints)
The feasible region is determined by these inequalities. The corner points of the feasible region are found by the intersection of the boundary lines:
- Point $A$: Intersection of $x = 2$ and $y = 4$,so $A = (2, 4)$.
- Point $B$: Intersection of $x = 2$ and $x + y = 5$,so $B = (2, 3)$.
- Point $C$: Intersection of $x + y = 5$ and $y = 0$,so $C = (5, 0)$.
Now,evaluate the objective function $z = 8x + 5y$ at these corner points:
- At $A(2, 4): z = 8(2) + 5(4) = 16 + 20 = 36$.
- At $B(2, 3): z = 8(2) + 5(3) = 16 + 15 = 31$.
- At $C(5, 0): z = 8(5) + 5(0) = 40 + 0 = 40$.
The minimum value of $z$ is $31$.
Therefore,the minimum cost is $₹ 31$.

(A) The constraints are $2x \geq 4$ (or $x \geq 2$),$y \leq 3$,$x+y \leq 8$,and $x, y \geq 0$.
The feasible region is bounded by the lines $x=2, y=3, x+y=8$ and the $x$-axis $(y=0)$.
The corner points of the feasible region are determined by the intersection of these lines:
$A(2, 0)$ (intersection of $x=2$ and $y=0$)
$B(8, 0)$ (intersection of $x+y=8$ and $y=0$)
$C(5, 3)$ (intersection of $x+y=8$ and $y=3$)
$D(2, 3)$ (intersection of $x=2$ and $y=3$)
Now,we evaluate $Z=100x+70y$ at each corner point:
At $A(2, 0): Z = 100(2) + 70(0) = 200$
At $B(8, 0): Z = 100(8) + 70(0) = 800$
At $C(5, 3): Z = 100(5) + 70(3) = 500 + 210 = 710$
At $D(2, 3): Z = 100(2) + 70(3) = 200 + 210 = 410$
The maximum value of $Z$ is $800$ at point $B(8, 0)$.

(B) The corner points of the feasible region are $O(0,0)$,$A(6,0)$,$B(6,4)$,$C(3,7)$,and $D(0,5)$.
We evaluate the objective function $z=4x+3y$ at each corner point:
At $O(0,0)$,$z=4(0)+3(0)=0$
At $A(6,0)$,$z=4(6)+3(0)=24$
At $B(6,4)$,$z=4(6)+3(4)=24+12=36$
At $C(3,7)$,$z=4(3)+3(7)=12+21=33$
At $D(0,5)$,$z=4(0)+3(5)=15$
Comparing these values,the maximum value of $z$ is $36$.

(C) The shaded region is bounded by three lines and the axes in the first quadrant:
$1$. The line passing through $(0, 3)$ and $(4, 0)$ is given by $\frac{x}{4} + \frac{y}{3} = 1$,which simplifies to $3x + 4y = 12$. Since the shaded region is towards the origin side,the inequality is $3x + 4y \leq 12$.
$2$. The line passing through $(0, 0)$ and $(3, 3)$ is $y = x$,which is $y - x = 0$. The shaded region is above this line,so the inequality is $y \geq x$ or $y - x \geq 0$.
$3$. The horizontal line passing through $(0, 3)$ is $y = 3$. The shaded region is below this line,so the inequality is $y \leq 3$.
$4$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
Combining these,the inequalities are $3x + 4y \leq 12, y - x \geq 0, y \leq 3, x, y \geq 0$.

(D) To find the feasible region,we analyze the given inequations in the $1^{\text{st}}$ quadrant $(x \geq 0, y \geq 0)$:
$1$. For $x-2y \leq 2$: The line passes through $(2, 0)$ and $(0, -1)$. Testing $(0, 0)$,$0-0 \leq 2$ is true,so the region is towards the origin side.
$2$. For $5x+2y \geq 10$: The line passes through $(2, 0)$ and $(0, 5)$. Testing $(0, 0)$,$0+0 \geq 10$ is false,so the region is away from the origin side.
$3$. For $4x+5y \leq 20$: The line passes through $(5, 0)$ and $(0, 4)$. Testing $(0, 0)$,$0+0 \leq 20$ is true,so the region is towards the origin side.
Combining these,the feasible region is the triangular area bounded by these lines in the $1^{\text{st}}$ quadrant,which corresponds to Fig. $3$.

(B) To determine the linear constraints for the shaded region,we analyze the position of the region relative to each line:
$1$. For the line $2x + y = 2$,the shaded region lies on the side away from the origin (non-origin side). Testing the point $(1, 1)$,we get $2(1) + 1 = 3 \geq 2$. Thus,the constraint is $2x + y \geq 2$.
$2$. For the line $x - y = 1$,the shaded region lies on the side containing the origin. Testing the point $(0, 0)$,we get $0 - 0 = 0 \leq 1$. Thus,the constraint is $x - y \leq 1$.
$3$. For the line $x + 2y = 8$,the shaded region lies on the side containing the origin. Testing the point $(0, 0)$,we get $0 + 2(0) = 0 \leq 8$. Thus,the constraint is $x + 2y \leq 8$.
Therefore,the required linear constraints are $2x + y \geq 2, x - y \leq 1, x + 2y \leq 8$.

(B) The objective function is $z = 3x + 5y$.
The constraints are $3x + 2y \leq 18$,$x \leq 4$,$y \leq 6$,and $x, y \geq 0$.
The feasible region is a polygon with corner points $O(0, 0)$,$A(4, 0)$,$B(4, 3)$,$C(2, 6)$,and $D(0, 6)$.
We evaluate $z$ at each corner point:
At $O(0, 0)$: $z = 3(0) + 5(0) = 0$
At $A(4, 0)$: $z = 3(4) + 5(0) = 12$
At $B(4, 3)$: $z = 3(4) + 5(3) = 12 + 15 = 27$
At $C(2, 6)$: $z = 3(2) + 5(6) = 6 + 30 = 36$
At $D(0, 6)$: $z = 3(0) + 5(6) = 30$
The maximum value of $z$ is $36$ at the point $(2, 6)$.

(C) The corner points of the given feasible region $OCDBO$ are $O(0, 0)$,$C(10, 10)$,$D(10, 20)$,and $B(0, 25)$.
We evaluate the objective function $z = 3x + 4y$ at each corner point:
$1$. At $O(0, 0)$: $z = 3(0) + 4(0) = 0$
$2$. At $C(10, 10)$: $z = 3(10) + 4(10) = 30 + 40 = 70$
$3$. At $D(10, 20)$: $z = 3(10) + 4(20) = 30 + 80 = 110$
$4$. At $B(0, 25)$: $z = 3(0) + 4(25) = 0 + 100 = 100$
Comparing the values $0, 70, 110,$ and $100$,the maximum value of $z$ is $110$.

(B) The feasible region is determined by the constraints $x+y \leq 20, y \geq 5, x \leq 10, x \geq 0, y \geq 0$.
The corner points of the feasible region are $A(0, 5), B(10, 5), C(10, 10),$ and $D(0, 20)$.
We evaluate the objective function $z = 7x + 8y$ at each corner point:
At $A(0, 5): z = 7(0) + 8(5) = 40$
At $B(10, 5): z = 7(10) + 8(5) = 70 + 40 = 110$
At $C(10, 10): z = 7(10) + 8(10) = 70 + 80 = 150$
At $D(0, 20): z = 7(0) + 8(20) = 160$
The maximum value of $z$ is $160$.

(B) $1$. Identify the lines bounding the region: The lines are $3x + 4y = 12$ (intercepts $(4,0)$ and $(0,3)$),$4x + 7y = 28$ (intercepts $(7,0)$ and $(0,4)$),and $y = 1$.
$2$. Analyze the inequalities:
- For the line $3x + 4y = 12$,the shaded region is away from the origin,so the inequality is $3x + 4y \geq 12$.
- For the line $4x + 7y = 28$,the shaded region is towards the origin,so the inequality is $4x + 7y \leq 28$.
- For the line $y = 1$,the shaded region is above the line,so the inequality is $y \geq 1$.
- Since the region is in the first quadrant,we have $x \geq 0$ and $y \geq 0$.
$3$. Combining these,the system of inequalities is $3x + 4y \geq 12, 4x + 7y \leq 28, y \geq 1, x \geq 0, y \geq 0$.

(C) The feasible region is a polygon with vertices $(0,0)$,$(3,0)$,$(3,2)$,$(2,3)$,and $(0,3)$.
We evaluate the objective function $z=10x+25y$ at each vertex:
At $(0,0)$,$z=10(0)+25(0)=0$
At $(3,0)$,$z=10(3)+25(0)=30$
At $(3,2)$,$z=10(3)+25(2)=30+50=80$
At $(2,3)$,$z=10(2)+25(3)=20+75=95$
At $(0,3)$,$z=10(0)+25(3)=75$
Comparing these values,the maximum value of $z$ is $95$.

(A) To find the feasible region,we analyze the given inequalities:
$1$. $4x + 3y \leq 60$: This represents the region on or below the line passing through $(15, 0)$ and $(0, 20)$.
$2$. $y \geq 2x$: This represents the region on or above the line passing through $(0, 0)$ and $(3, 6)$.
$3$. $x \geq 3$: This represents the region to the right of the vertical line $x = 3$.
$4$. $x, y \geq 0$: This represents the first quadrant.
By testing a point $(4, 10)$ which lies in the $S_2$ region:
- $4(4) + 3(10) = 16 + 30 = 46 \leq 60$ (True)
- $10 \geq 2(4) = 8$ (True)
- $4 \geq 3$ (True)
- $4, 10 \geq 0$ (True)
Since all conditions are satisfied,the solution set is represented by the $S_2$ region.

(D) To determine the linear constraints,we analyze the boundary lines of the shaded region:
$1$. The line passing through $(3, 0)$ and $(0, 2)$ is $2x + 3y = 6$. Since the shaded region is away from the origin,the constraint is $2x + 3y \geq 6$.
$2$. The line passing through $(0, 3)$ and $(6, 0)$ is $3x + 6y = 18$. Since the shaded region is towards the origin,the constraint is $3x + 6y \leq 18$.
$3$. The line passing through $(3, 0)$ and $(0, -1)$ is $x - 3y = 3$. Since the shaded region is towards the origin,the constraint is $x - 3y \leq 3$.
$4$. The line passing through $(0, 1)$ and $(2, 2)$ is $-x + 2y = 2$. Since the shaded region is towards the origin,the constraint is $-x + 2y \leq 2$.
$5$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
Thus,the correct constraints are $2x + 3y \geq 6, 3x + 6y \leq 18, x - 3y \leq 3, -x + 2y \leq 2, x \geq 0, y \geq 0$.

(D) The given system of in-equations is:
$x+y \leq 70$
$x+2y \leq 100$
$2x+y \leq 120$
$x \geq 0, y \geq 0$
Since all the inequalities $x+y \leq 70$,$x+2y \leq 100$,and $2x+y \leq 120$ are satisfied by the origin $(0,0)$ (as $0+0 \leq 70$,$0+0 \leq 100$,and $0+0 \leq 120$ are all true),the feasible region must lie towards the origin in the first quadrant.
By analyzing the lines and the constraints $x \geq 0, y \geq 0$,the common shaded region that satisfies all these conditions is represented by Fig. $3$.

(C) The feasible region is determined by the constraints $2x+3y \leq 12$,$2x+y \leq 8$,$x \geq 0$,and $y \geq 0$. The corner points of the feasible region are $(0, 0)$,$(4, 0)$,$(3, 2)$,and $(0, 4)$.
We evaluate the objective function $z=4x+5y$ at these corner points:
At $(0, 0)$: $z = 4(0) + 5(0) = 0$
At $(4, 0)$: $z = 4(4) + 5(0) = 16$
At $(3, 2)$: $z = 4(3) + 5(2) = 12 + 10 = 22$
At $(0, 4)$: $z = 4(0) + 5(4) = 20$
The maximum value of $z$ is $22$ at the point $(3, 2)$.

(D) The feasible region is bounded by the corner points $(0, 80)$,$(14, 33)$,and $(80, 0)$.
We evaluate the objective function $Z = 4x + 6y$ at each corner point:
$1$. At $(0, 80)$: $Z = 4(0) + 6(80) = 480$
$2$. At $(14, 33)$: $Z = 4(14) + 6(33) = 56 + 198 = 254$
$3$. At $(80, 0)$: $Z = 4(80) + 6(0) = 320$
Comparing these values,the minimum value of $Z$ is $254$,which occurs at the point $(14, 33)$.

(A) To find the maximum value of $Z = 5x + 2y$,we identify the feasible region defined by the constraints:
$1$. $2x - y \geq 2$
$2$. $x + 2y \leq 8$
$3$. $x, y \geq 0$
The boundary lines are $2x - y = 2$ and $x + 2y = 8$.
- For $2x - y = 2$,the intercepts are $(1, 0)$ and $(0, -2)$.
- For $x + 2y = 8$,the intercepts are $(8, 0)$ and $(0, 4)$.
To find the intersection point of $2x - y = 2$ and $x + 2y = 8$:
Multiply the first equation by $2$: $4x - 2y = 4$.
Adding this to $x + 2y = 8$,we get $5x = 12$,so $x = 2.4$.
Substituting $x = 2.4$ into $2x - y = 2$: $2(2.4) - y = 2 \implies 4.8 - y = 2 \implies y = 2.8$.
The vertices of the feasible region are $(1, 0)$,$(8, 0)$,and $(2.4, 2.8)$.
Now,evaluate $Z = 5x + 2y$ at these vertices:
- At $(1, 0)$: $Z = 5(1) + 2(0) = 5$
- At $(8, 0)$: $Z = 5(8) + 2(0) = 40$
- At $(2.4, 2.8)$: $Z = 5(2.4) + 2(2.8) = 12 + 5.6 = 17.6$
Comparing these values,the maximum value is $40$ at $(8, 0)$.

(D) To find the maximum value of $z=50x+15y$,we evaluate the objective function at the corner points of the feasible region.
The corner points of the feasible region are $(0, 0)$,$(20, 0)$,$(10, 50)$,and $(0, 60)$.
Evaluating $z$ at each point:
$1$. At $(0, 0)$: $z = 50(0) + 15(0) = 0$
$2$. At $(20, 0)$: $z = 50(20) + 15(0) = 1000$
$3$. At $(10, 50)$: $z = 50(10) + 15(50) = 500 + 750 = 1250$
$4$. At $(0, 60)$: $z = 50(0) + 15(60) = 900$
The maximum value is $1250$,which occurs at the point $(10, 50)$.

(D) To determine the linear constraints for the shaded region,we analyze each boundary line:
$1$. The line $x+2y=6$ has the shaded region on the side away from the origin (since the point $(7, 6)$ satisfies $7+12=19 \geq 6$). Thus,the constraint is $x+2y \geq 6$.
$2$. The line $5x+3y=15$ has the shaded region on the side away from the origin (since the point $(7, 6)$ satisfies $35+18=53 \geq 15$). Thus,the constraint is $5x+3y \geq 15$.
$3$. The vertical line $x=7$ has the shaded region to its left,so $x \leq 7$.
$4$. The horizontal line $y=6$ has the shaded region below it,so $y \leq 6$.
$5$. The region is in the first quadrant,so $x, y \geq 0$.
Combining these,the constraints are $x+2y \geq 6, 5x+3y \geq 15, x \leq 7, y \leq 6, x, y \geq 0$.

(B) The feasible region is bounded by the lines $x+y=5$,$4x+3y=24$,$x=0$,and $y=0$.
To find the corner points,we identify the vertices of the shaded region:
$1$. The intersection of $x=0$ and $4x+3y=24$ gives $(0, 8)$.
$2$. The intersection of $x=0$ and $x+y=5$ gives $(0, 5)$.
$3$. The intersection of $x+y=5$ and $4x+3y=24$ is found by solving the system: $y=5-x \implies 4x+3(5-x)=24 \implies x=9, y=-4$ (Not in the first quadrant).
Looking at the graph,the vertices are $(0, 5)$,$(0, 8)$,and the intersection of $4x+3y=24$ with the $x$-axis $(6, 0)$ and $x+y=5$ with the $x$-axis $(5, 0)$.
Evaluating $Z=2x+y$ at the corner points:
At $(0, 5)$,$Z = 2(0) + 5 = 5$.
At $(0, 8)$,$Z = 2(0) + 8 = 8$.
At $(5, 0)$,$Z = 2(5) + 0 = 10$.
At $(6, 0)$,$Z = 2(6) + 0 = 12$.
The maximum value is $12$ at the point $(6, 0)$.

(C) To find the point in the feasible region,we must check which point satisfies all the given inequalities:
$1$) $x+y \leq 3$
$2$) $2x+5y \geq 10$
$3$) $x \geq 0, y \geq 0$
Let us test the given options:
For option $(A)$ $(2,2)$: $2+2 = 4 \not\leq 3$. (False)
For option $(B)$ $(4,2)$: $4+2 = 6 \not\leq 3$. (False)
For option $(C)$ $(1,2)$: $1+2 = 3 \leq 3$ (True),$2(1)+5(2) = 2+10 = 12 \geq 10$ (True),and $1 \geq 0, 2 \geq 0$ (True). (True)
For option $(D)$ $(2,1)$: $2(2)+5(1) = 4+5 = 9 \not\geq 10$. (False)
Thus,the point $(1,2)$ satisfies all the inequalities and lies in the feasible region.

(B) To determine the linear constraints,we analyze the half-planes defined by each line for the shaded region:
$1$. For the line $3x + 4y = 18$,the origin $(0,0)$ satisfies $3(0) + 4(0) = 0 \leq 18$. Since the shaded region contains the origin relative to this line,the constraint is $3x + 4y \leq 18$.
$2$. For the line $2x + 3y = 3$,the origin $(0,0)$ gives $2(0) + 3(0) = 0 \leq 3$. Since the shaded region is on the side away from the origin,the constraint is $2x + 3y \geq 3$.
$3$. For the line $x - 6y = 3$,the origin $(0,0)$ gives $0 - 0 = 0 \leq 3$. Since the shaded region is on the side containing the origin,the constraint is $x - 6y \leq 3$.
$4$. For the line $-x + 2y = 2$,the origin $(0,0)$ gives $0 + 0 = 0 \leq 2$. Since the shaded region is on the side containing the origin,the constraint is $-x + 2y \leq 2$.
$5$. The region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.
Thus,the correct set of constraints is $3x + 4y \leq 18, 2x + 3y \geq 3, x - 6y \leq 3, -x + 2y \leq 2, x \geq 0, y \geq 0$.

(A) $1$. The shaded region is bounded by the lines $x=0$ ($Y$-axis),$y=0$ ($X$-axis),$x=5$,$y=3$,and the line passing through $(0,0)$ and $(3,3)$.
$2$. The line passing through $(0,0)$ and $(3,3)$ has the equation $y=x$,which can be written as $x-y=0$. Since the shaded region lies below this line,the constraint is $x-y \geq 0$.
$3$. The vertical line $x=5$ bounds the region on the right,so $x \leq 5$.
$4$. The horizontal line $y=3$ bounds the region on the top,so $y \leq 3$.
$5$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
$6$. Combining these,the constraints are $x, y \geq 0, x-y \geq 0, x \leq 5, y \leq 3$.

(B) The constraints are $2x + y \geq 7$,$2x + 3y \leq 15$,$y \leq 3$,$x \geq 0$,and $y \geq 0$.
From the graph,the vertices of the feasible region are $A(3.5, 0)$,$B(7.5, 0)$,$C(3, 3)$,and $D(2, 3)$.
We evaluate the objective function $z = 4x + 5y$ at these vertices:
$z(A) = 4(3.5) + 5(0) = 14 + 0 = 14$
$z(B) = 4(7.5) + 5(0) = 30 + 0 = 30$
$z(C) = 4(3) + 5(3) = 12 + 15 = 27$
$z(D) = 4(2) + 5(3) = 8 + 15 = 23$
The minimum value is $14$,which occurs at point $A(3.5, 0)$. Since point $A$ lies on the $X$-axis,the minimum value occurs on the $X$-axis.

(B) The feasible region is determined by the constraints $0 \leq x \leq 3$,$0 \leq y \leq 3$,and $x + y \leq 5$.
The vertices of the feasible region are:
$O(0, 0)$,$A(3, 0)$,$B(3, 2)$,$C(2, 3)$,and $D(0, 3)$.
We evaluate the objective function $Z = 10x + 25y$ at each vertex:

Vertex	$Z = 10x + 25y$
$O(0, 0)$	$10(0) + 25(0) = 0$
$A(3, 0)$	$10(3) + 25(0) = 30$
$B(3, 2)$	$10(3) + 25(2) = 30 + 50 = 80$
$C(2, 3)$	$10(2) + 25(3) = 20 + 75 = 95$
$D(0, 3)$	$10(0) + 25(3) = 75$

Comparing the values,the maximum value of $Z$ is $95$,which occurs at the point $(2, 3)$.

(A) The feasible region is determined by the constraints $x+y \leq 5$,$2x+y \geq 4$,$x \geq 0$,and $y \geq 0$.
From the graph,the vertices of the feasible region are $A(0, 4)$,$B(2, 0)$,$C(5, 0)$,and $D(0, 5)$.
We evaluate the objective function $z = 2x + 3y$ at each vertex:
At $A(0, 4)$: $z = 2(0) + 3(4) = 12$.
At $B(2, 0)$: $z = 2(2) + 3(0) = 4$.
At $C(5, 0)$: $z = 2(5) + 3(0) = 10$.
At $D(0, 5)$: $z = 2(0) + 3(5) = 15$.
The maximum value of $z$ is $15$ at point $D(0, 5)$.