Question

Solve the following problem :

Maximize Z = 5x₁ + 6x₂ Subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x ≥ 0, x₂ ≥ 0

Answer 1

To find the graphical solution, construct the table as follows:

Inequation	equation	Double intercept form	Points (x1, x2)	Region
2x1 + 3x2 ≤ 18	2x1 + 3x2 = 18	`x_1/(9) + x_2/(6)` = 1	A (9, 0) B (0, 6)	2(0) + 3(0) ≤ 18 ∴ 0 ≤ 18 ∴ Origin-side
2x1 + x2 ≤ 12	2x1 + x2 = 12	`x_1/(6) + x_2/(12)` = 1	C (6, 0) D (0, 12)	2(0) + 1(0) ≤ 12 ∴ 0 ≤ 12 ∴ Origin-side
x1 ≥ 0	x1 = 0	–	–	R.H.S. of Y-axis
x2 ≥ 0	x2 ≥ 0	–	–	above X-axis

The shaded portion OBEC is the feasible region.
Whose vertices are O (0, 0), B (0, 6), E, C (6, 0)
E is the point of intersection of the lines
2x₁ + x₂ = 12       ...(i)
and 2x₁ + 3x₂ = 18  ...(ii)
∴ By (i) – (ii), we get
2x₁ + x₂   = 12
2x₁ + 3x₂ = 18
 –     –       –    
      –2x₂ = – 6
∴  x₂ = `(-6)/(-2)` = 3
Substituting x2 = 3 in (i), we get
2x₁ + 3 = 12
∴ 2x₁ = 12 –  3
∴ 2x₁ = 9
∴ x₁ = `(9)/(2)` = 4.5
∴ E (4.5,3)
Here, the objective function is Z = 5x₁ + 6x₂
Now, we will find maximum value of Z as follows:

Feasible points	The value of Z = 5x1 + 6x2
O (0, 0)	Z = 5(0) + 6(0) = 0
B (0, 6)	Z = 5(0) + 6(6) = 36
E (4.5, 3)	Z = 5(4.5) + 6(3) = 22.5 + 18 = 40.5
E (4.5, 3)	Z = 5(6) + 6(0) = 30

∴ Z has maximum value 40.5 at E(4.5, 3)
∴ Z is maximum, when x₁ = 4.5, x₂ = 3.