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

Given a Linear Programming Problem (LPP) with the objective function to maximize $z = 3x + 4y$ subject to constraints.
The constraints defining the feasible region are:
$x + y \leq 40$, $x + 2y \leq 60$, and $x, y \geq 0$.
To find the optimal solution, we first determine the vertices of the feasible region:
1. Intersection of $x + y = 40$ and $x + 2y = 60$:
Solving $x + y = 40$ for $x$ gives $x = 40 - y$. Substituting into the second equation:
$(40 - y) + 2y = 60 \Rightarrow y = 20$. Substituting $y=20$ back into $x = 40 - y$ gives $x = 20$. Thus, the intersection point is (20, 20).
2. Intersection with the y-axis ($x=0$):
From $x + y \leq 40$, if $x=0$, then $y \leq 40$.
From $x + 2y \leq 60$, if $x=0$, then $2y \leq 60 \Rightarrow y \leq 30$.
The feasible points on the y-axis are (0, 0) and (0, 30).
3. Intersection with the x-axis ($y=0$):
From $x + y \leq 40$, if $y=0$, then $x \leq 40$.
From $x + 2y \leq 60$, if $y=0$, then $x \leq 60$.
The most restrictive condition is $x \leq 40$, so the feasible point on the x-axis is (40, 0).
Now, we evaluate the objective function $z = 3x + 4y$ at each corner point:
At (0,0): $z = 3(0) + 4(0) = 0$
At (0,30): $z = 3(0) + 4(30) = 120$
At (20,20): $z = 3(20) + 4(20) = 60 + 80 = 140$
At (40,0): $z = 3(40) + 4(0) = 120$
Revisiting the constraints: The intersection of $x + 2y = 60$ and $y=0$ yields $x=60$, resulting in the point (60,0). However, this point is outside the feasible region defined by $x+y \leq 40$. The vertices of the feasible region are (0,0), (0,30), (40,0), and (20,20).
Comparing the z-values, the maximum value of $z$ is 140, which occurs at the point (20, 20).
Final Answer: Correct Answer: (B) 140