Question

In a LPP, the maximum value of $z = 3x + 4y$ subject to the constraints $x + y \leq 40$, $x + 2y \leq 60$, $x, y \geq 0$ is

• A. 120
• B. 140
• C. 150
• D. 130

Hint:

Always evaluate the objective function at each vertex of the feasible region to find the optimal value in an LPP.

Answer 1

The Correct Option is C

We are given an LPP (Linear Programming Problem) where the objective is to maximize $z = 3x + 4y$ under constraints.
To solve it, we identify the feasible region determined by the constraints:
$x + y \leq 40$, $x + 2y \leq 60$, and $x, y \geq 0$.
Find points of intersection (vertices) of the feasible region:
1. Intersection of $x + y = 40$ and $x + 2y = 60$:
Solving, $x + y = 40 \Rightarrow x = 40 - y$, substitute in second:
$40 - y + 2y = 60 \Rightarrow y = 20, x = 20$ $\Rightarrow$ point (20, 20)
2. $x = 0$: then from $x + y \leq 40$ $\Rightarrow$ $y \leq 40$ and $x + 2y \leq 60$ $\Rightarrow$ $2y \leq 60$ $\Rightarrow$ $y \leq 30$
So, point (0, 0) and (0, 30) lie on the boundary.
3. $y = 0$: from $x + y \leq 40$ $\Rightarrow$ $x \leq 40$ and $x + 2y \leq 60$ $\Rightarrow$ $x \leq 60$
So, point (40, 0) is valid.
Now evaluate $z = 3x + 4y$ at all corner points:
At (0,0): $z = 0$
At (0,30): $z = 3(0) + 4(30) = 120$
At (20,20): $z = 3(20) + 4(20) = 60 + 80 = 140$
At (40,0): $z = 3(40) + 4(0) = 120$
Wait: let's double-check. One corner is missed.
Intersection of $x + 2y = 60$ and $x + y = 40$ already gave (20,20)
Another key point: where $x + 2y = 60$ intersects $y=0$ $\Rightarrow$ $x = 60$ $\Rightarrow$ (60,0)
But $x + y = 40$ $\Rightarrow$ (40,0) is more limiting.
So max is at (20,20) $\Rightarrow$ $z = 140$
Final correction: Correct Answer: (B) 140