A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B. Each product is processed on two machines M1 and M2. One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2, whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.
Let x units of product A and y units of product B were manufactured.
Number of products cannot be negative.
Therefore, \[x, y \geq 0\]
According to question, the given information can be tabulated as
The constraints are
\[x + y \leq 300\] \[2x + y \leq 360\]
Total profit = Z = \[5x + 3y\] which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z= \[5x + 3y\]
\[x + y \leq 300\] .
\[2x + y \leq 360\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
x + y = 300, 2x + y = 360, x = 0 and y = 0
Region represented by x + y ≤ 300:
The line x + y = 300 meets the coordinate axes at A1(300, 0) and B1(0, 300) respectively. By joining these points we obtain the line
x + y = 30. Clearly (0,0) satisfies the x + y = 30. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 300.
Region represented by 2x + y ≤ 360:
The line 2x + y = 360 meets the coordinate axes at C1(180, 0) and D1(0, 360) respectively. By joining these points we obtain the line
2x + y = 360. Clearly (0,0) satisfies the inequation 2x + y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 360.
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 ≤ 300, 2x + y ≤ 360, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B1(0, 300), E1(60, 240) and C1(180, 0).
The values of Z at these corner points are as follows
|Corner point||Z= 5x + 3y|
The maximum value of Z is Rs 1020 which is attained at B1\[\left( 60, 240 \right)\]
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