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प्रश्न
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
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उत्तर
We have to maximize Z = 60x + 15y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 50, 3x + y = 90, x = 0 and y = 0
Region represented by x + y ≤ 50:
The line x + y = 50 meets the coordinate axes at A(50,0) and B(0,50) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation x + y ≤ 50. So,the region containing the origin represents the solution set of the inequation x + y ≤ 50.
Region represented by 3x + y ≤ 90:
The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively. By joining these points we obtain the line 3x + y = 90.
Clearly (0,0) satisfies the inequation 3x + y ≤ 90. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 90.
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, x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), C(30, 0),
The values of Z at these corner points are as follows.
| Corner point | Z = 60x + 15y |
| O(0, 0) | 60 × 0 + 15 × 0 = 0 |
| C(30, 0) | 60 × 30 + 15 × 0 = 1800 |
|
\[E\left( 20, 30 \right)\]
|
60 × 20 + 15 × 30 =1650 |
| B(0, 50) | 60 × 0 + 15 × 50 = 750 |
Therefore, the maximum value of Z is \[1800 \text{ at the point } \left( 30, 0 \right)\] Hence, x = 30 and y = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 1800.
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