A merchant plans to sell two types of personal computers desktop model and a portable model that will cost Rs25000 and Rs40000 respectively. He estimates that the total monthly demand for computers will not exceed 250 units. Determine the number of units of each type of computer that the merchant should stock to get maximum profit if he does not want to invest more than Rs70lakhs and if his profit on the desktop model is Rs4500 and on portable model is Rs5000.
A merchant plans to sell two types of personal computers desktop model and a portable model that will cost Rs25000 and Rs40000 respectively. He estimates that the total monthly demand for computers will not exceed 250 units. Determine the number of units of each type of computer that the merchant should stock to get maximum profit if he does not want to invest more than Rs70lakhs and if his profit on the desktop model is Rs4500 and on portable model is Rs5000.
Solution and Explanation
Let the merchant stock,
x desktop models and y portable models.
Therefore, x≥0 and y≥0
The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000.
However, the merchant can invest a maximum of Rs 70 lakhs.
∴25000x+40000y≤7000000
5x+8y≤1400
The monthly demand for computers will not exceed 250 units.
∴x+y≤250
The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.
Total profit, Z=4500x+5000y
Thus, the mathematical formulation of the given problem is Maximum Z=4500x+5000y...(1)
Subject to the constraints, 5x+8y≤1400....(2)
x+y≤250...(3)
x,y≥0...(4)
The feasible region determined by the system of constraints is as follows.
The corner points are A(250,0), B(200,50) and C(0,175).
The value of Z at these corner points is as follows.
| Corner point | Z=4500x5000y | |
| A(250,0) | 1125000 | |
| B(200,50) | 1150000 | \(\rightarrow\)Maximum |
| C(0,175) | 875000 |
The maximum value of Z is 1150000 at (200,50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs 1150000.
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Notes on Linear Programming Problem
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.