Question

Minimize Z = 2x + 3y subject to constraints

x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + y ≥ 6	2x + y ≥ 7	x + 4y ≥ 8
Corresponding equation (of line)	x + y = 6	2x + y = 7	x + 4y = 8
Intersection of line with X-axis	(6, 0)	`(7/2, 0)`	(8, 0)
Intersection of line with Y-axis	(0, 6)	(0, 7)	(0, 2)
Region	Non-origin side	Non-origin side	Non-origin side

Shaded portion XABCDY is the feasible region, whose vertices are A(8, 0), B, C and D(0, 7).

B is the point of intersection of x + 4y = 8 and x + y = 6.

Solving the above equations, we get

x = `16/3`, y = `2/3`

∴ B ≡ `(16/3, 2/3)`

C is the point of intersection of 2x + y = 7 and x + y = 6.

Solving the above equations, we get

x = 1, y = 5

∴ C ≡ (1, 5)

Here, the objective function is Z = 2x + 3y

∴ Z at A(8, 0) = 2(8) + 3(0) = 16

Z at B`(16/3, 2/3)`

= `2(16/3) + 3(2/3)`

= `32/3 + 6/3`

= `38/3`

Z at C(1, 5) = 2(1) + 3(5) = 17

Z at D(0, 7) = 2(0) + 3(7) = 21

∴ Z has minimum value `38/3` at x = `16/3` and y = `2/3`.