Question

Minimise Z=3x+5y
such that x+3y≥3,x+y≥2,x,y≥0

Answer 1

The feasible region determined by the system of constraints,
x+3y≥3, x+y≥2, and x,y≥0 is as follows.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(3,0), B($\frac{3}{2}$,$\frac{1}{2}$)and C(0,2).

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

As this feasible region is unbounded,
therefore,7 may or may not be the minimum value of Z.

For this we draw the graph of the inequality,3x+5y<7, and check whether the resulting half-plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 3x+5y<7

Therefore, the minimum value of Z is 7 at ($\frac{3}{2}$,$\frac{1}{2}$).