Question

Maximize z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 5y ≤ 26	5x + 3y ≤ 30
Corresponding equation (of line)	3x + 5y = 26	5x + 3y = 30
Intersection of line with X-axis	`(26/3, 0)`	(6, 0)
Intersection of line with Y-axis	`(0, 26/5)`	(0, 10)
Region	Origin side	Origin side

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(6, 0), B and C `(0, 26/5)`.

B is the point of intersection of the lines 5x + 3y = 30 and 3x + 5y = 26. Solving the above equations, we get

x = `9/2`, y = `5/2`

∴ B ≡ `(9/2, 5/2)` ≡ `(4.5, 2.5)`

Here, the objective function is Z = 7x + 11y

∴ Z at O(0, 0) = 7(0) + 11(0) = 0

Z at A(6, 0) = 7(6) + 11(0) = 42

Z at B`(9/2, 5/2) = 7(9/2) + 11(5/2)`

= `(63 + 55)/2`

= 59

Z at C`(0, 26/5) - 7(0) + 11(26/5)`

= `286/5`

= 57.2

∴ Z has maximum value 59 at B`(9/2, 5/2)`.

i.e. Z has maximum value 59 when x = 4.5 and y = 2.5