Question

Solve the following problem :

Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0

Answer 1

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

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

The shaded portion Y' DE is the feasible region.
Whose vertices are D(0, 3) and E
E is the point of intersection of lines
x – y = 1          …(i)
x + y = 3         …(ii)
∴ By (i) + (ii), we get
     x – y = 1
    x + y = 3 
   2x      = 4

∴ x = `(4)/(2)` = 2
Substituting x = 2 in (i), we get
2 – y = 1
∴ y  = 1
∴ E(2, 1)
Here, the objective function is Z = 2x + 3y
Now, we will find minimum value of Z as follows:

Feasible points	The value of Z = 2x + 3y
D(0, 3)	Z = 2(0) + 3(3) = 9
E(2, 1)	Z = 2(2) + 3(1) = 4 + 3 = 7

∴ Z has minimum value 7 at E(2, 1)
∴ Z is minimum, when x = 2, y = 1.