Question

Solve the following LPP:

Maximize z = 4x + 2y subject to 3x + y ≤ 27, x + y ≤ 21, x ≥ 0, y ≥ 0.

Answer 1

First we draw the lines AB and CD whose equations are 3x + y = 27 and x + y = 21 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + y = 27	A(9, 0)	B(0, 27)	≤	origin side of line AB
CD	x + y = 21	C(21, 0)	O(0, 21)	≤	origin side of line CD

The feasible region is OAPDO which is shaded region in the graph. The vertices of the feasible region are O(0, 0), A (9, 0), P and D (0, 21). P is the point of intersection of lines

3x + y = 27      ....(1)

and x + y = 21      ....(2)

On substracting, we get 2x = 6 

∴ x = 3

Substituting x = 3 in equation (1), we get

9 + y = 27

∴ y = 18

∴ P ≡ (3, 18)

The values of the objective function z = 4x + 2y at these vertices are

z(O) = 4(0) + 2(0) = 0 + 0 = 0

z(a) = 4(9) + 2(0) = 36 + 0 = 36

z(P) = 4(3) + 2(18) = 12 + 36 = 48

z(D) = 4(0) + 2(21) = 0 + 42 = 42

∴ 2 has minimum value 48 when x = 3, y = 18.