Question

Choose the correct alternative :

The maximum value of z = 5x + 3y. subject to the constraints

Options

• 235

• `(235)/(9)`

• `(235)/(19)`

• `(235)/(3)`

Answer 1

Z = 5x + 3y
The inequalities are 3x + 5y ≤ 15, 5x + 2y ≤ 10
Consider lines L₁ and L₂ where
L₁ : 3x + 5y = 15, 5x + 2y = 10
For line L₁, Plot A (0, 3) and B (5, 0)
For line L₂, plot P(0, 5) and Q(2, 0)

Solving both line, we get x = `(20)/(19), y = (45)/(19)`
The coordinates of the origin O (0, 0) satisfies both the inequalities.

∴ The required region is on the origin side of both the lines L₁ and L₂.
As x ≥ 0, y ≥ 0; the feasible region is in the 1^st quadrant.
OQRAO is the required feasible region.
At O (0, 0), Z = 0
At Q (2, 0), Z = 5(2) + 0 = 10

At R `(20/19, 45/19) , z = 5(20/19) + 3(45/19) = (235)/(19)`.

At A (0, 3), Z = 0 + 3(3) = 9

The maximum value of Z is `(235)/(19)` and it occurs at point R`(20/19, 45/19)`.