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प्रश्न
A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each of product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.
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उत्तर
Let x units of product A and y units of product B were manufactured.
Number of units cannot be negative.
Therefore, \[x, y \geq 0\]
According to question, the given information can be tabulated as:
| Selling price(Rs) | Manufacturing time(hrs) | |
| Product A(x) | 200 | 0.5 |
| Product B(y) | 300 | 1 |
Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.
Further, it is given that there is a permanent order for 14 units of product A and 16 units of product B.
Therefore, the constraints are
\[200x + 300y \geq 10000\]
\[0 . 5x + y \leq 40\]
\[x \geq 14\]
\[y \geq 16\]
If the profit on each of product A is Rs 20 and on product B is Rs 30.Therefore, profit gained on x units of product A and y units of product 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
\[2x + 3y \geq 100\]
\[x + 2y \leq 80\]
\[x \geq 14\]
\[y \geq 16\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
2x + 3y = 100, x + 2y = 80, x = 14, y = 16, x = 0 and y = 0.
Region represented by 2x + 3y ≥ 100:
The line 2x + 3y = 100 meets the coordinate axes at A1(50, 0) and \[B_1 \left( 0, \frac{100}{3} \right)\] respectively. By joining these points we obtain the line 2x + 3y = 100 .
Clearly (0,0) does not satisfies the 2x + 3y = 100. So, the region which does not contain the origin represents the solution set of the inequation 2x + 3y ≥ 100.
Region represented by x + 2y ≤ 80:
The line x + 2y = 80 meets the coordinate axes at C1(80, 0) and D1(0, 40) respectively. By joining these points we obtain the line
x + 2y = 80. Clearly (0,0) satisfies the inequation x + 2y ≤ 80. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 80.
Region represented by x ≥ 14
x = 14 is the line passes through (14, 0) and is parallel to the Y axis.The region to the right of the line x = 14 will satisfy the inequation.
Region represented by y ≥ 16
y = 16 is the line passes through (0, 16) and is parallel to the X axis.The region above the line y = 16 will satisfy the inequation.
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 2x + 3y ≥ 100, x + 2y ≤ 80, x≥ 14, y ≥ 16, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E1(26, 16), F1(48, 16), G1(14, 33) and H1(14, 24)
The values of Z at these corner points are as follows
| Corner point | Z= 20x + 30y |
| E1 | 1000 |
| F1 | 1440 |
| G1 | 1270 |
| H1 | 1000 |
The maximum value of Z is Rs 1440 which is attained at F1
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