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Question
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?
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Solution
Let x units of product A and y units of product B were manufactured.
Clearly, \[x \geq 0, y \geq 0\]
It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.The two products are produced in a common production process, which has a total capacity of 500 man-hours.
\[5x + 3y \leq 500\]
The maximum number of unit of A that can be sold is 70 and that for B is 125.
\[x \leq 70\]
\[y \leq 125\]
If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B. Therefore, profit x units of product A and y units of product B is Rs 20x and Rs 15y respectively.
Total profit = Z = \[20x + 15y\]
The mathematical formulation of the given problem is
Max Z = \[20x + 15y\]
subject to
\[5x + 3y \leq 500\]
\[x \leq 70\]
\[y \leq 125\]
First we will convert inequations into equations as follows:
5x + 3y = 500, x = 70, y = 125, x = 0 and y = 0
Region represented by 5x + 3y ≤ 500:
The line 5x + 3y = 500 meets the coordinate axes at A1(100, 0) and \[B_1 \left( 0, \frac{500}{3} \right)\] respectively. By joining these points we obtain the line 5x + 3y = 500. Clearly (0,0) satisfies the 5x + 3y = 500. So, the region which contains the origin represents the solution set of the inequation 5x + 3y ≤ 500.
Region represented by x ≤ 70:
The line x = 70 is the line passes through C1(70, 0) and is parallel to Y axis. The region to the left of the line x = 70 will satisfy the inequation x ≤ 70.
Region represented by y ≤ 125:
The line y = 125 is the line passes through D1(0, 125) and is parallel to X axis. The region below the the line y = 125 will satisfy the inequation y ≤ 125.
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 5x + 3y ≤ 500, x ≤ 70, y ≤ 125, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D1 \[\left( 0, 125 \right)\] , E1(25, 125), F1(70, 50) and C1(70, 0).The values of Z at the corner points are
| Corner points | Z = \[20x + 15y\] |
| O | 0 |
| D1 | 1875 |
| E1 | 2375 |
| F1 | 2150 |
| C1 | 1400 |
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