Question

Solve the following LPP by graphical method:

Maximize: z = 7x + 11y, subject to:

3x + 4y ≤ 24, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.

Answer 1

First, we draw the lines AB and CD, whose equations are 3x + 4y = 24 and 5x + 3y = 30, respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + 4y = 24	A (8, 0)	B (0, 6)	≤	origin side of the line AB
CD	5x + 3y = 30	C (6, 0)	D(0, 10)	≤	origin side of the line CD

The feasible region is OCPBO, which is shaded in the figure.

The vertices of the feasible region are O (0, 0), C (6, 0), P and B (0, 6)

The vertex P is the point of intersection of the lines

3x + 4y = 24    ...(1)

5x + 3y = 30    ...(2)

Multiplying equation (1) by 3 and equation (2) by 4, we get

9x + 12y = 72 and

20x + 12y = 120

On subtracting, we get

11 x = 48

∴ `x = 48/11`

Substituting `x = 48/11` in equation (2), we get

`5(48/11) + 3y = 30`

∴ `3y = 30 - 240/11`

= `90/11`

∴ `y = 30/11`

∴ P is `(48/11, 30/11)`

The values of the objective function z = 7x + 11y at these corner points are

z(O) = 7(0) + 11(0) = 0 + 0 = 0

z(C) = 7(6) + 11(0) = 42 + 0 = 42

z(P) = `7(48/11) + 11(30/11) `

= `336/11 + 330/11`

= `666/11`

= 60.54

z(B) = 7(0) + 11(6)

= 0 + 66

= 66

∴ z has maximum value 66, when x = 0 and y = 6