Question

Solve the following LPP:

Maximize z = 4x₁ + 3x₂ subject to
3x₁ + x₂ ≤ 15, 3x₁ + 4x₂ ≤ 24, x₁ ≥ 0, x₂ ≥ 0. 

Answer 1

We first draw the lines AB and CD whose equations are 3x₁ + x₂ = 15 and 3x₁ + 4x₂ = 24 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x1 + x2 = 15	A(5, 0)	B(0,15)	≤	origin side of the line AB
CD	3x1 + 4x2 = 24	C(8, 0)	D(0, 6)	≤	origin side of the line CD

The feasible region is OAPDO which is shaded in the graph.

The Vertices of the feasible region are O(0, 0), A(5, 0), P and D(0, 6).

P is the point of intersection of lines.

3x₁ + 4x₂ = 24       ....(1)

and 3x₁ + x₂ = 15      ....(2)

On subtracting, we get

3x₂ = 9    ∴ x₂ = 3

Substituting x₂ = 3 in (2), we get

3x₁ + 3 = 15 

∴ 3x₁ = 12

∴ x₁ = 4

∴ P is (4, 3)

The values of objective function z = 4x₁ + 3x₂ at these vertices are

z(O) = 4(0) + 3(0) = 0 + 0 = 0

z(a) = 4(5) + 3(0) = 20 + 0 = 20

z(P) = 4(4) + 3(3) = 16 + 9 = 25

z(D) = 4(0) + 3(6) = 0 + 18 = 18

∴ z has maximum value 25 when x = 4 and y = 3.