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Question
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
Options
235
`(235)/(9)`
`(235)/(19)`
`(235)/(3)`
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Solution
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 1st 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)`.
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