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State whether the following statement is True or False:
The maximum value of Z = 5x + 3y subjected to constraints 3x + y ≤ 12, 2x + 3y ≤ 18, 0 ≤ x, y is 20
Options
True
False
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Solution
False
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Inequation | Equation | X intercept | Y intercept | Region |
2x + 3y ≤ 18 | 2x + 3y = 18 | (9, 0) | (0, ___) | Towards origin |
2x + y ≤ 10 | 2x + y = 10 | ( ___, 0) | (0, 10) | Towards origin |
x ≥ 0, y ≥ 0 | x = 0, y = 0 | X axis | Y axis | ______ |
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P( ___, ___ ), C(5, 0)
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Point | Coordinates | Z = 9x + 13y | Values | Remark |
O | (0, 0) | 9(0) + 13(0) | 0 | |
A | (0, 6) | 9(0) + 13(6) | ______ | |
P | ( ___,___ ) | 9( ___ ) + 13( ___ ) | ______ | ______ |
C | (5, 0) | 9(5) + 13(0) | ______ |
∴ Z is maximum at __( ___, ___ ) with the value ___.
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Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
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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 | 1^{st} 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 ___
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Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
x, 0, y ≥ 0 ......(iv)
At which point minimum value of Z is attained.