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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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Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
| x, y ≥ 0 | x = 0, y = 0 | x = 0 | y = 0 | 1st quadrant |
∵ Origin has not satisfied the inequations.
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In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
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The coordinates of C are obtained by solving equations
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Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
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