Question

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.

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

The Vertices of the feasible region are 0(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.