Question

Minimize z = 7x + y subjected to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.

Answer 1

To draw the feasible region, construct a table as follows:

Inequality	5x + y ≥ 5	x + y ≥ 3
Corresponding equation (of line)	5x + y = 5	x + y = 3
Intersection of line with X-axis	(1, 0)	(3, 0)
Intersection of line with Y-axis	(0, 5)	(0, 3)
Region	Non-origin side	Non-origin side

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B, and C(0, 5).

B is the point of intersection of the lines x + y = 3 and 5x + y = 5.

Solving the above equations, we get x = `1/2`, y = `5/2`

∴ B ≡ `(1/2, 5/2)`

Here, the objective function is Z = 7x + y

∴ Z at A(3, 0) = 7(3) + 0 = 21

Z at B`(1/2, 5/2) = 7(1/2) + 5/2 = 7/2 + 5/2` = 6

Z at C(0, 5) = 7(0) + 5 = 5

∴ Z has a minimum value of 5 at C(0, 5).

∴ Z has a minimum value of 5 when x = 0 and y = 5.