Question

Solve the L.P.P. graphically:

Minimize: z = 5x + 2у,

Subject to: 5x + y ≥ 10, x + y ≥ 6, 

                  x ≥ 0, y ≥ 0

Answer 1

Given:

Minimize: z = 5x + 2у

Subject to:

5x + y ≥ 10

x + y ≥ 6

x ≥ 0, y ≥ 0

Step 1: Find corner points of feasible region

From the graph, feasible region lies above both lines in first quadrant.

Corner points are:

1. Intersection with y-axis of 5x + y = 10

(0, 10)

2. Intersection of both lines:

Solve:

5x + y = 10

x + y = 6

Subtract:

4x = 4

⇒ x = 1

y = 5

So point: (1, 5)

3. Intersection with x-axis of x + y = 6:

(6, 0)

Step 2: Calculate z at corner points

z = 5x + 2y

At (0, 10): z = 20

At (1, 5): z = 5(1) + 2(5) = 5 + 10 = 15

At (6, 0): z = 30

Minimum value of z = 15

At point: (x, y) = (1, 5)