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Question
Solve the following LPP by using graphical method.
Maximize : Z = 6x + 4y
Subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
Also find maximum value of Z.
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Solution
| Inequalities | x ≤ 2 | x+y≤3 | -2x+y≤1 |
| Equalities | x=2 | x+y=3 | -2x+y=1 |
| Intercept form |
`x/2=1` |
`x/3+y/3=1` |
`x/(-1/2)+y/1=1` |
| Origin Test | 0≤2 | 0+0≤3 | -2(0)+0≤1 |
| True Origin Side | True Origin Side | True Origin Side |
Shaded portion OABC is the feasible region, Where O(0,0) A(2, 0) D(0, 1), B(2, 1)
For C :
x + y = 3
– 2x + y = 1
– – –
----------------------
3x = 2
∴ x = 2/3
2/3+y=3 i.e y=7/3
`∴ c(2/3,7/3)`
Z = 6x + 4y
Z at O(0, 0) = 6(0) + 4(0) = 0
Z at A(2, 0) = 6(2) + 4(0) = 12
Z at B(2, 1) = 6(2) + 4(1) = 16
Z at `c(2/3,7/3)=6(2/3)+(7/3)4=40/3`
Z at D(0,1) = 6(0) + 4(1) = 4 Thus, Z is maximized at B(2, 1) and its maximum value is 16.
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