Question

Solve the following linear programming problem graphically:

Maximize Z = 8x + 9у

Subject to the constraints:

2x + 3y ≤ 6

3x − 2y < 6

y ≤ 1

x > 0, y ≥ 0

Answer 1

Maximize Z = 8x + 9y

Subject to the constraints.

2x + 3y ≤ 6

3x − 2y ≤ 6

y < 1 and x ≥ 0, y ≥ 0

Convert into equations,

2x + 3y = 6

3x − 2y = 6

y = 1

2x + 3y = 6

3x − 2y = 6

x	0	3
y	2	0

and

x	0	2
y	-3	0

on solving equations (iii) and (i)

2x + 3y = 6 and y = 1

2x + 3 = 6

2x = 6 − 3

2x = 3

x = `3/2`

B`(3/2, 1)`

On solving equations (i) and (ii),

`3((6 - 3y)/2) - 2y = 6`

`9 - (9y)/2 - 2y = 6`

`9 - 6 = (9y)/2 + 2y`

`3 = (13y)/2`

3 × 2 = 13y

6 = 13y

∴ y = `6/13`

and x = `(6 - 3(6/13))/2`

= `((6 - 3)(6/13))/2`

= `(6 - 18/13)/2`

= `((78 - 18)/13)/2`

= `(60/13)/(2)`

= `60/(13 xx 2)`

`x = 30/13`

Now, `(30/13, 6/13)`

According to graph feasible region is OABCD.

Corner Point	Maximize Z = 8x + 9у
O(0, 0)	Z = 8 × 0 + 9 × 0 = 0
A(0, 1)	Z = 8 × 0 + 9 × 1 = 9
B`(3/2, 1)`	`Z = 8 xx 3/2 + 9 xx 1 = 21`
C`(30/13, 6/13)`	`Z = 8 xx 30/13 + 9 xx 6/13 = 294/13`
D(2, 0)	Z = 8 × 2 + 9 × 0 = 16

Hence, the maximum value of Z is `294/13` at C`(30/13, 6/13)`