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प्रश्न
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
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उत्तर
First, we will convert the given inequations into equations, we obtain the following equations:
3x + y = 12, 2x + 5y = 34, x = 0 and y = 0
Region represented by 3x + y ≤ 12:
The line 3x + y = 12 meets the coordinate axes at \[A\left( 4, 0 \right)\] and \[B\left( 0, 12 \right)\] respectively. By joining these points we obtain the line 3x + y = 12.
Clearly (0,0) satisfies the inequation 3x + y ≤ 12. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 12 .
Region represented by 2x + 5y ≤ 34:
The line 2x + 5y = 34 meets the coordinate axes at \[C\left( 17, 0 \right)\] and \[D\left( 0, \frac{34}{5} \right)\]
respectively. By joining these points we obtain the line 2x + 5y ≤ 34.
Clearly (0,0) satisfies the inequation 2x + 5y ≤ 34. So,the region containing the origin represents the solution set of the inequation 2x + 5y ≤ 34.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), \[A\left( 4, 0 \right)\], \[E\left( 2, 6 \right)\] and \[D\left( 0, \frac{34}{5} \right)\] .
The values of Z at these corner points are as follows:
| Corner point | Z = 10x + 6y |
| O(0, 0) | 10 × 0 + 6 × 0 = 0 |
|
\[A\left( 4, 0 \right)\]
|
10× 4 + 6 × 0 = 40 |
|
\[E\left( 2, 6 \right)\]
|
10 × 2 + 6 × 6 = 56 |
|
\[D\left( 0, \frac{34}{5} \right)\]
|
10 × 0 + 6 × \[\frac{34}{5}\] = \[\frac{204}{3}\]
|
We see that the maximum value of the objective function Z is 56 which is at \[E\left( 2, 6 \right)\] that means at x = 2 and y = 6.
Thus, the optimal value of Z is 56.
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