Question

Consider the LPP Maximise; Z = 5x + 3y Subject to; 3x + 5y < 15, 5x + 2y < 10x, y > 0

(i) Draw the feasible region. 

(ii) Find the corner points of the feasible region.

(iii) Find the corner at which Z attains its maximum.

Answer 1

In the figure, the shaded region OABC is the feasible region. Here the region is bounded.

O(0,0) , A(2,0) ,\(B(\frac{20}{19},\frac{45}{19})\),C(0,3)

Given; Z = 5x + 3y 

Corner points	Value of Z
O	Z = 0
A	Z = 5(2)+3(0) = 10
B	Z = 5\(\frac{20}{19} + 3 (\frac{45}{19})=\frac{235}{19}\)
C	Z = 5(0)+3(3) = 9

Since maximum value of Z occurs at B, the solution is Z = 10,9