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Question
Solve the following problem:
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0
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Solution
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x1, x2) | Region |
| 3x1 + x2 ≤ 15 | 3x1 + x2 = 15 | `x_1/(5) + x_2/(15)` = 1 | A (5, 0) B (0, 15) |
3(0) + 0 ≤ 15 ∴ 0 ≤ 15 ∴ origin side |
| 3x1 + 4x2 ≤ 24 | 3x1 + 4x2 = 24 | `x_1/8 + x_2/(6)` = 1 | C (8, 0) D (0, 6) |
3(0 + 4(0) ≤ 24 ∴ 0 ≤ 24 ∴ origin side |
| x1 ≥ 0 | x1 = 0 | – | – | R.H.S. of Y-axis |
| x2 ≥ 0 | x2 = 0 | – | – | above X-axis |
Shaded portion ODEA is the feasible region.
Whose vertices are O (0, 0), D (0, 6), E, A (5, 0)
E is the point of intersection of the lines
3x1 + x2 = 15 …(i)
and 3x1 + 4x2 = 24 …(ii)
∴ By (i) – (ii), we get
3x1 + x2 = 15
3x1 + 4x2 = 24
– – –
–3x2 = – 9
∴ x2 = `(-9)/(-3)`
∴ x2 = 3
Substituting x2 = 3 in i, we get
3x1 + 3 = 15
∴ 3x1 = 15 – 3
∴ 3x1 = 12
∴ x1 = `(12)/(3)` = 4
∴ E (4, 3)
Here, the objective function is Z = 4x1 + 3x2
Now, we will find maximum value of Z as follows:
| Feasible points | The value of Z = 4x1 + 3x2 |
| O (0, 0) | Z = 4(0) + 3(0) = 0 |
| D (0, 6) | Z = 4(0) + 3(0) = 0 |
| E (4, 3) | Z = 4(4) + 3(3) = 16 + 9 = 25 |
| A (5, 0) | Z = 4(4) + 3(3) = 16 + 9 = 25 |
∴ Z has maximum value 25 at E (4, 3)
∴ Z is maximum, when x1 = 4, x2 = 3.
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