Question

Solve the following LPP Graphically;

Maximise; Z = 3x + 5y

Subject to constraints;

x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0

Answer 1

In the figure the shaded region ABC is the fesible region. Here the region is unbouded.

The corner points are A(3, 0), B(\(\frac{3}{2}\),\(\frac{1}{2}\)), C(0, 2)

Given; Z = 3x + 5y

Corner points	Value of Z
A	Z = 3(3) + 5(0) = 9
B	Z = 3(\(\frac{3}{2}\)) + 5(\(\frac{1}{2}\)) = 7
C	Z = 3(0) + 5(2) = 10

 Form the table, minumum value of Z is 7 at B(\(\frac{3}{2}\),\(\frac{1}{2}\)). The feasible region is unbounded, so consider the inequality 3x + 5y < 7. Clearly the feasible region has no common points with 3x + 5y < 7, Thus minimum value of Z occurs at B, the solution is Z = 7.