Question

Solve the following Linear Programming Problem graphically.

Maximise Z = 5x + 2y subject to:

x – 2y ≤ 2,

3x + 2y ≤ 12,

– 3x + 2y ≤ 3,

x ≥ 0, y ≥ 0

Answer 1

Given L.P.P. is

Max. Z = 5x + 2y subject to

x – 2y ≤ 2,

3x + 2y ≤ 12,

– 3x + 2y ≤ 3,

and x ≥ 0, y ≥ 0

First, we draw graphs of lines

x – 2y = 2        ...(1)

3x + 2y = 12    ...(2)

– 3x + 2y = 3   ...(3)

By (1) x – 2y = 2

x	2	0
y	0	– 1

A(2, 0) and B(0, – 1)

By (2) 3x + 2y = 12

x	4	0
y	0	6

C(4, 0) and D(0, 6)

By (3) – 3x + 2y = 3

x	– 1	0
y	0	3/2

`E(-1, 0), F(0, 3/2)`

Now solving (1) and (2), we get

x = `7/2` and y = `3/4`

Point of intersection `P(7/2, 3/4)`

Solving (2) and (3), we get

x = `3/2`, y = `15/4`

Point of intersection `Q(3/2, 15/4)`

x ≥ 0, y ≥ 0 shows that the shaded region lies in first quadrant.

For x – 2y ≤ 2, take origin (0, 0), then 0 – 0 ≤ 2 which is true.

∴ Shaded the region in which origin lies.

Similarly, we shade the region for various situations.

Hence OAPQFO is the required feasible region.

The corresponding Z values can be evaluated as follows:

Vertices of shaded region	Z = 5z + 2y
O(0, 0)	Z = 5 × 0 + 2 × 0 = 0
A(2, 0)	Z = 10 + 0 = 10
`P(7/2, 3/4)`	Z = `35/2 + 3/2` = 19 (maximum)
`Q(3/2, 15/4)`	Z = `15/2 + 15/2` = 15
`F(0, 3/2)`	Z = 0 + 3 = 3

Hence, the maximum value of Z is 19 at the point `P(7/2, 3/4)`.