Vardaan Learning Institute

Ans: Milling constraint: $2x + y \le 40$. Grinding constraint: $x + 2y \le 50$.

Ans: Vitamin A: $2x + y \ge 8$. Vitamin C: $x + 2y \ge 10$.

Ans: $x + y \le 250$.

Ans: $25000x + 40000y \le 7000000 \Rightarrow 5x + 8y \le 1400$.

Ans: Since depot A has 20 trucks total, the remaining trucks sent to E will be $20 - (x + y)$.

Ans: Since factory C needs 15 trucks and $x$ come from A, depot B must send $15 - x$ trucks to factory C.

Ans: Maximize $Z = 4500x + 5000y$.

Ans: $x \ge 0, y \ge 0, z \ge 0$.

Ans: Maximum value is $16$ at the point $(0, 4)$.

Ans: Corner points: $(0,0)$, $(4,0)$, $(2,3)$, and $(0,4)$. Minimum value is $-12$ at $(4,0)$.

Ans: Evaluating $Z$: $Z(0,0)=0$, $Z(4,0)=-12$, $Z(2,3)=6$, $Z(0,4)=16$. Maximum value is $16$ at $(0,4)$.

Ans: Solving $5x+2y=10$ and $3x+5y=15$. Multiply eq1 by 5 and eq2 by 2: $25x+10y=50$ and $6x+10y=30$. Subtracting gives $19x = 20 \Rightarrow x = \frac{20}{19}$. Substitute $x$: $y = \frac{45}{19}$. The corner point is $(\frac{20}{19}, \frac{45}{19})$.

Ans: Corner points: $(0,0)$, $(30,0)$, $(20,30)$, and $(0,50)$. Max value is $120$ at $(30,0)$.

Ans: $(0,0)$, $(3,0)$, $(3,4)$, and $(0,4)$.

Ans: $Z(0,0)=0$, $Z(3,0)=6$, $Z(0,4)=20$, $Z(3,4)=26$. The maximum value is $26$.

Ans: Multiply $x+y=5$ by 2: $2x+2y=10$. Subtracting from $2x+3y=12$ gives $y=2$. Hence $x=3$. The point is $(3, 2)$.

Ans: Corner points are $(10,0)$, $(60,0)$, $(0,20)$, and $(0,10)$. $Z(10,0)=30$, $Z(60,0)=180$, $Z(0,20)=180$, $Z(0,10)=90$. The maximum value is $180$ occurring at both $(60,0)$ and $(0,20)$.

Ans: The minimum value is $30$ at the point $(10,0)$.

Ans: Corner points of the triangle are $(4,3)$, $(0,5)$, and $(0,6)$. $Z(4,3) = 2300$, $Z(0,5) = 2500$, $Z(0,6) = 3000$. The minimum value is $2300$ at $(4,3)$.

Ans: Intersection of $x+3y=3$ and $x+y=2$ is $(1.5, 0.5)$. Corner points are $(3,0)$, $(1.5, 0.5)$, and $(0,2)$. $Z(3,0)=9$, $Z(1.5, 0.5)=7$, $Z(0,2)=10$. The minimum value is $7$.

Ans: To confirm it's the minimum, we must graph the open half-plane $3x + 5y < 7$ as a dotted line.

Ans: No. Since the dotted line $3x+5y < 7$ does not overlap with the feasible region, $7$ is the absolute minimum.

Ans: The corner points are $(0,3)$ and $(6,0)$. $Z(0,3) = 6$ and $Z(6,0) = 6$. The minimum value is $6$ (which occurs at all points on the line segment joining them).

Ans: No. The feasible region extends infinitely. The test $ -x + 2y > 1$ shares points with the feasible region, meaning $Z$ can increase indefinitely without bound.

Ans: The maximum value does not exist because the objective function can be made infinitely large within the unbounded feasible region.

Ans: $3p + 4q = 4p + 3q \Rightarrow p = q$.

Ans: There is no feasible region (the constraints contradict each other). Hence, no maximum value exists.

Ans: Infinitely many optimal solutions (every point on the line segment joining $(0, 20)$ and $(15, 15)$ is optimal).

Ans: It guarantees that both the absolute maximum and absolute minimum values of the objective function must exist, and they occur at the corner points (vertices) of the bounded region.

Ans: No. The region required is $y \ge x+1$ AND $y \le x$. It is geometrically impossible for a value to be simultaneously strictly greater than $x+1$ and less than $x$. Therefore, the LPP has an infeasible solution.