Question

Solve the following LPP graphically ;

Maximise; Z = 60x +15y

subject to constraints;

x+y ≤ 50, 3x+y ≤ 90, x ≥ 0, y ≥ 0.

Answer 1

1. In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(30, 0), B(20, 30), C(0, 50).

Given; Z = 60x + 15y

Corner points	Value of Z
O	Z = 0
A	Z = 60(30) +15(0) = 1800
B	Z = 60(20) + 15(30) = 16500
C	Z = 60(0) + 15(50)=750

Since maximum value of Z occurs at A, the solution is Z = 1800, (30, 0).