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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)`
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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Minimize Z = 4x + 5y
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.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
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 ___
