Question

Maximise Z=5x+3y
Subject to 3x+5y≤15,5x+2y≤10,x≥0,y≥0.

Answer 1

The feasible region determined by the system of constraints,
3x+5y≤15, 5x+2y≤10,x≥,y≥0 are as follows.

The corner points of the feasible region are O(0,0),A(2,0),B(0,3)and C($\frac{20}{19}$,$\frac{45}{19}$).

The value of Z at these corner points are as follows.

Therefore,the maximum value of Z is $\frac{235}{29}$ at the point ($\frac{20}{19}$,$\frac{45}{19}$).