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Question
State whether the following statement is True or False:
The graphical solution set of the inequations 0 ≤ y, x ≥ 0 lies in second quadrant
Options
True
False
Solution
False
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Inequation | Equation | X intercept | Y intercept | Region |
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2x + y ≤ 10 | 2x + y = 10 | ( ___, 0) | (0, 10) | Towards origin |
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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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5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
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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.
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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 ___