A company manufactures two types of novelty souvenirs made of polywood.Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and building.3 hours 20 minutes are available for cutting and 4 hours of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many types of souvenirs of each type should the company manufacture in order to maximize the profit?
A company manufactures two types of novelty souvenirs made of polywood.Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and building.3 hours 20 minutes are available for cutting and 4 hours of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many types of souvenirs of each type should the company manufacture in order to maximize the profit?
Solution and Explanation
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Therefore, \(x≥0,y≥0\)
The given information can be compiled in a table as follows.
| Type A | Type B | Availability | |
| Cutting(min) | 5 | 8 | 3×60+20=200 |
| Assembling(min) | 10 | 8 | 4×60=240 |
The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6.
Therefore, the constraints are
\(5x+8y≤200\)
\(10x+8y≤240\)
i.e., \(5x+4y≤120\)
Total profit, \(Z=5x+6y\)
The mathematical formulation of the given problem is Maximize \(Z=5x+6y\) ……...(1)
Subject to the constraints,
\(5x+8y≤200\) …....(2)
\(5x+4y≤120\) …....(3)
\(x,y≥0\) …....(4)
The feasible region determined by the system of constraints is as follows.
The corner points are A(24,0), B(8,20) and C(0,25).
The value of Z at these corner points is as follows.
| Corner point | Z=5x+6y |
| A(24,0) | 120 |
| B(8,20) | 160 (Max) |
| C(0,25) | 150 |
The maximum value of Z is 200 at (8,20).
Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs. 160.
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Concepts Used:
Linear Programming Problems
The Linear Programming Problems (LPP) is a problem that is concerned with finding the optimal value of the given linear function. The optimal value can be either maximum value or minimum value. Here, the given linear function is considered an objective function. The objective function can contain several variables, which are subjected to the conditions and it has to satisfy the set of linear inequalities called linear constraints.
Linear Programming Simplex Method
Step 1: Establish a given problem. (i.e.,) write the inequality constraints and objective function.
Step 2: Convert the given inequalities to equations by adding the slack variable to each inequality expression.
Step 3: Create the initial simplex tableau. Write the objective function at the bottom row. Here, each inequality constraint appears in its own row. Now, we can represent the problem in the form of an augmented matrix, which is called the initial simplex tableau.
Step 4: Identify the greatest negative entry in the bottom row, which helps to identify the pivot column. The greatest negative entry in the bottom row defines the largest coefficient in the objective function, which will help us to increase the value of the objective function as fastest as possible.
Step 5: Compute the quotients. To calculate the quotient, we need to divide the entries in the far right column by the entries in the first column, excluding the bottom row. The smallest quotient identifies the row. The row identified in this step and the element identified in the step will be taken as the pivot element.
Step 6: Carry out pivoting to make all other entries in column is zero.
Step 7: If there are no negative entries in the bottom row, end the process. Otherwise, start from step 4.
Step 8: Finally, determine the solution associated with the final simplex tableau.