Question

Minimize z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0 show that the minimum value of z occurs at more than two points

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 2y ≥ 3	x + 4y ≥ 4	3x + y ≥ 3
Corresponding equation (of line)	x + 2y = 3	x + 4y = 4	3x + y = 3
Intersection of line with X-axis	(3, 0)	(4, 0)	(1, 0)
Intersection of line with Y-axis	`(0, 3/2)`	(0, 1)	(0, 3)
Region	Non-origin side	Non-origin side	Non-origin side

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Shaded portion XABCDY is the feasible region, whose vertices are A(4, 0), B, C and D(0, 3).

B is the point of intersection of the lines x + 4y = 4 and x + 2y = 3.

Solving the above equations, we get

x = 2, y = `1/2`

∴ B ≡ `(2, 1/2)`

C is the point of intersection of the lines x + 2y = 3 and 3x + y = 3.

Solving the above equations, we get

x = `3/5`, y = `6/5`

∴ C ≡ `(3/5, 6/5)`

Here, the objective function is Z = 6x + 21y

∴ Z at A(4, 0) = 6(4) + 21(0) = 24

Z at B`(2, 1/2) = 6(2) + 21(1/2)`

= `12 + 21/2`

= `45/2`

= 22.5

Z at C`(3/5, 6/5) = 6(3/5) + 21(6/5)`

= `18/5 + 126/5`

= `144/5`

= 28.8

Z at D(0, 3) = 6(0) + 21(3) = 63

∴ Z has minimum value 22.5 at B`(2, 1/2)`.

∴ Z has minimum value 22.5 when x = 2 and y = 0.5.