Question

Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0

Answer 1

Inequation	Equation	Double intercept form	Points (x, y)	Region
x + 2y ≤ 40	x + 2y = 40	`x/(40) + y/(20)` = 1	A (40, 0) B (0, 20)	0 + 2(0) ≤ 40 ∴ 0 ≤ 40 ∴ origin-side
3x + 2y ≤ 60	3x + 2y = 60	`x/(20) + y/(30)` = 1	C (20, 0) D (0, 30)	3(0) + 2(0) ≤ 60 ∴ 0 ≤  60 ∴ origin-side
x ≥ 0	x = 0	–	–	R.H.S. of Y-axis
y ≥ 0	y = 0	–	–	Above X-axis

 

Shaded portion OBEC is the feasible region

Whose vertices are O (0, 0), B(0, 20), E and C (20, 0) 

E is the point of intersection of lines

x + 2y = 40        …(i)

3x + 2y = 60      …(ii)

∴ By (i) – (ii), we get

 x + 2y = 40
3x + 2y = 60
–    –      –       
     –2x  = –20

∴ x = `(-20)/(-2)`

∴ x = 10

Substituting x = 10 in (i), we get 

10 + 2y = 40

∴ 2y = 40 – 10

∴ 2y = 30

∴ y = `(30)/(2)` = 15

∴ E = (10, 15)

Here, the objective function is Z = 60x + 50y 

Now, we will find maximum value of Z as follows:

Feasible points	The value of Z = 60x + 50y
O (0, 0)	Z = 60(0) + 50(0) = 0
B (0, 20)	Z = 60(0) + 50(20) = 1000
E (10, 15)	Z = 60(10) + 50(15) = 600 + 750 = 1350
C (20, 0)	Z = 60(20) + 50(0) = 1200

∴ Z has maximum value 1350 at E (10, 15)

∴ Z is maximum, when x = 10, y = 15.