Question

Maximise Z = 20x + 10y subject to constraints

Subject to x + 2y ≥ 40

3x + y ≥ 30

4x + 3y ≥ 60

x,y ≥ 0

Answer 1

On changing the inequality into equation, we have

x + 2y = 40, 3x + y = 30, 4x + 3y = 60

We first draw the graph of given line

The shaded region EAQPE is the feasible region.

The vertices of the feasible region are E (15,0), A(40,0), Q (4,18) and P(6,12)

Now, the value of the objective function are as :

Given, z = 20x + 10y

At E (15,0) , z = 20 × 15 + 10 × 0 = 300

At A(40,0), z = 20 × 40 + 10 × 0 = 800

At Q(4,18), z = 20 × 4 + 10 × 18 = 260

At P(6,12), z = 20 × 6 + 10 × 12 = 240

Obviosly, z is minimum at P(6,12)

Hence, x = 6, y = 12 is optimal solution of the given LPP and the optimal value of z is 240.