Question

Solve the following problem :

Minimize 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

The shaded portion CHB is the feasible region.
Whose vertices are C(21, 0), H and B(0, 27)
H is the point of intersection of lines
3x + y = 27 …(i)
x + y = 21  …(ii)
∴ By (i) – (ii), we get
3 x + y = 27
   x + y = 21
–    –     –      
2x        = 6

∴ x = `(6)/(2)`  = 3
Substituting x = 3 in (ii), we get
3 + y = 21
∴ y = 18
∴ H (3, 18)
Here, the objective function is Z = 4x + 2y
Now, we will find minimum value of Z as follows:

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

∴ Z has minimum value 48 at H (3, 18)
∴ Z is minimum, when x = 3, y = 18