Question

Solve the following problem:

Maximize Z = 4x₁ + 3x₂ Subject to 3x₁ + x₂ ≤ 15, 3x₁ + 4x₂ ≤ 24, x₁ ≥ 0, x₂ ≥ 0

Answer 1

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

Inequation	Equation	Double intercept form	Points (x1, x2)	Region
3x1 + x2 ≤ 15	3x1 + x2 = 15	`x_1/(5) + x_2/(15)` = 1	A (5, 0) B (0, 15)	3(0) + 0 ≤ 15 ∴ 0 ≤ 15 ∴ origin side
3x1 + 4x2 ≤  24	3x1 + 4x2 = 24	`x_1/8 + x_2/(6)` = 1	C (8, 0) D (0, 6)	3(0 + 4(0) ≤ 24 ∴ 0 ≤ 24 ∴ origin side
x1 ≥ 0	x1 = 0	–	–	R.H.S. of Y-axis
x2 ≥ 0	x2 = 0	–	–	above X-axis

Shaded portion ODEA is the feasible region.
Whose vertices are O (0, 0), D (0, 6), E, A (5, 0)
E is the point of intersection of the lines
3x1 + x2 = 15               …(i)
and 3x1 + 4x2 = 24     …(ii)
∴ By (i) – (ii), we get
   3x1 +   x2 = 15
   3x1 + 4x2 = 24
    –    –        –      
          –3x2 = – 9

∴ x₂ = `(-9)/(-3)`

∴ x₂ = 3
Substituting x₂ = 3 in i, we get
3x₁ + 3 = 15
∴ 3x₁ = 15 – 3
∴ 3x₁ = 12
∴ x₁ = `(12)/(3)` = 4
∴ E (4, 3)
Here, the objective function is Z = 4x₁ + 3x₂
Now, we will find maximum value of Z as follows:

Feasible points	The value of Z = 4x1 + 3x2
O (0, 0)	Z = 4(0) + 3(0) = 0
D (0, 6)	Z = 4(0) + 3(0) = 0
E (4, 3)	Z = 4(4) + 3(3) = 16 + 9 = 25
A (5, 0)	Z = 4(4) + 3(3) = 16 + 9 = 25

∴ Z has maximum value 25 at E (4, 3)
∴ Z is maximum, when x₁ = 4, x₂ = 3.