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Question
Solve the following Linear Programming problem graphically:
Maximize: Z = 3x + 3.5y
Subject to constraints:
x + 2y ≥ 240,
3x + 1.5y ≥ 270,
1.5x + 2y ≤ 310,
x ≥ 0, y ≥ 0.
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Solution
Changing inequations to equations
x + 2y = 240
3x + 1.5y = 270
1.5x + 2y = 310
For equation (i)
| x | 0 | 240 |
| y | 120 | 0 |
For equation (ii)
| x | 0 | 90 |
| y | 180 | 0 |
For equation (iii)
| x | 0 | 206.7 |
| y | 155 | 0 |
For A:
(3x + 1.5y = 270) × 2
(1.5x + 2y = 310) × 3/2
`\implies` 6x + 3y = 540
2.25x + 3y = 465
– – –
3.75x = 75
x = `75/3.75` = 20
y = `(310 - (1.5 xx 20))/2`
= `(310 - 30)/2`
= 140
A = (20, 140)
For B:
3x + 1.5y = 270
(x + 2y = 240) × 3
`\implies` 3x + 1.5y = 270
3x + 6y = 720
– – –
– 4.5y = – 450
`\implies` y = 100
x = 40
B = (40, 100)
For C:
x + 2y = 240
1.5x + 2y = 310
– – –
– 0.5x = – 70
x = 140
y = 50
C = (140, 50)
| Corner points | Max. Z = 3x + 3.5y |
| A(20, 140) | (3 × 20) + (3.5 × 140) = 550 |
| B(40, 100) | (3 × 40) + (3.5 × 100) = 470 |
| C(140, 50) | (3 × 140) + (50 × 3.5) = 595 `rightarrow` Max. |
Maximum value is 595 at (140, 50).
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