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प्रश्न
A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
उत्तर
Let x units of item A and y units of item B were manufactured.
Number of items cannot be negative.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as follows:
| Product | Motors | Transformers |
| A(x) | 3 | 4 |
| B(y) | 2 | 4 |
| Availability | 210 | 300 |
Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.
Therefore, the constraints are
\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]
A and B can make a profit of Rs 20 per unit of A and Rs 30 per unit of B.Therefore, profit gained from x units of item A and y units of item B is Rs 20x and Rs 30y respectively.
Total profit = Z = \[20x + 30y\] which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[20x + 30y\]
subject to
\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
3x + 2y = 210, 4x + 4y = 300, y = 65, x = 0 and y = 0
Region represented by 3x + 2y ≤ 210:
The line 3x + 2y = 210 meets the coordinate axes at A1(70, 0) and B1(0, 105) respectively. By joining these points we obtain the line 3x + 2y = 210. Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 210.
Region represented by 4x + 4y ≤ 300:
The line 4x + 4y = 300 meets the coordinate axes at C1(75, 0) and D1(0, 75) respectively. By joining these points we obtain the line
4x + 4y = 300.Clearly (0,0) satisfies the inequation 4x + 4y ≤ 300. So,the region which contains the origin represents the solution set of the inequation
4x + 4y ≤ 300.
y = 65 is the line passing through the point E1(0, 65) and is parallel to X axis.
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 + 2y ≤ 210, 4x + 4y ≤ 300, y ≤ 65, x ≥ 0, and y ≥ 0 are as follows
The corner points are O(0, 0), E1(0, 65), G1
The values of Z at these corner points are as follows
| Corner point | Z= 20x + 30y |
| O | 0 |
| E1 | 1950 |
| G1 | 2150 |
| F1 | 1650 |
| A1 | 1400 |
The maximum value of Z is 2150 which is attained at G1
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