Question

Solve the following problems by graphical method:

Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0 y ≥ 0

Answer 1

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

Inequation	Equation	Double intercept form	Points (x1,x2)	Points (x1,x2)
3x + y ≥ 27	3x + y = 27	`x/9+y/27=1`	A(9,0) B(0,27)	3(0) + 0 ≥ 27  0 ≥ 27 ∴ Non-origin side
x + y ≥ 21	x + y = 21	`x/21+y/21=1`	C(21,0) D(0,21)	(0) + 0 ≥ 21 ∴ 0 ≥ 21 ∴ Non-origin side
x ≥ 0	x = 0	-	-	R.H.S of Y-axis
y ≥ 0	y = 0	-	-	above X-axis

BHC is the unbounded feasible region with B = (0, 27) and C = (21,0)

Point H is the point of intersection of lines

3x + y = 27     ...(i)

x + y = 21     ...(ii)

Subtracting (ii) from (i),

   3x + y = 27
    x + y = 21
    –  –      –       
      2x  = 6

∴ x = 3

Put x =3 in (ii), 3 + 4 = 21

∴ y = 18

∴ H = (3,18)

Objective function, Z = 4x + 24

Feasible points	The value of Z = 4x + 2y
B (0,27)	Z = 4(0) + 2(27) = 54
H (3,18)	Z = 4(3) + 2(18) = 12 + 36 = 48
C (21,0)	Z = 4(21) + 2(0) = 84 + 0 = 84

∴  At H (3,18) the value of Z is minimum.

Hence Z has minimum value 48, when x = 3, y = 18.