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Question
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
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Solution
Let x units of first product and y units of second product be manufactured.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as follows:
| Product | Resource A | Resource B | Resource C |
| First(x) | 2 | 2 | 4 |
| Second(y) | 4 | 2 | 0 |
| Availability | 20 | 12 | 16 |
Therefore, the constraints are
\[2x + 4y \leq 20\]
\[2x + 2y \leq 12\]
\[4x + 0y \leq 16 \text{ or } 4x \leq 16\]
It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Therefore, profit gained from x units of first product and y units of second product is 2x monetary units and 4y monetary units respectively.
Total profit = Z = \[2x + 3y\] which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[2x + 3y\]
subject to
\[2x + 4y \leq 20\]
\[2x + 2y \leq 12\]
\[4x + 0y \leq 16 \text { or} 4x \leq 16\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows :
2x + 4y = 20, 2x + 2y = 12, 4x = 16, x = 0 and y = 0
Region represented by 2x + 4y ≤ 20:
The line 2x + 4y = 20 meets the coordinate axes at A1(10, 0) and B1(0, 5) respectively. By joining these points we obtain the line 2x + 4y = 20. Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 20.
Region represented by 2x + 2y ≤ 12:
The line 2x +2y =16 meets the coordinate axes at C1(6, 0) and D1(0, 6) respectively. By joining these points we obtain the line 2x + 2y = 12. Clearly (0,0) satisfies the inequation 2x + 2y ≤ 12. So, the region which contains the origin represents the solution set of the inequation 2x + 2y ≤ 12.
Region represented by 4x ≤ 16:
The line 4x =16 or x = 4 is the line passing through the point E1(4, 0) and is parallel to Y axis.The region to the left of the line x = 4 would satisfy the inequation 4x ≤ 16.
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 + 4y ≤ 20, 2x + 2y ≤ 12, 4x ≤ 16, x ≥ 0 and y ≥ 0 are as follows
The corner points are O(0, 0), B1(0, 5), G1 \[\left( 2, 4 \right)\] F1(4, 2) and E1(4, 0).
The values of Z at these corner points are as follows
| Corner point | Z= 2x + 3y |
| O | 0 |
| B1 | 15 |
| G1 | 16 |
| F1 | 14 |
| E1 | 8 |
The maximum value of Z is 16 which is attained at G1 \[\left( 2, 4 \right)\] Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufacture .
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