Question:

A factory manufactures two types of screws,A and B each type of screw requires the use of two machine,an automatic and a hand operated.It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A,while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screw B.Each machine is available for at the most 4 hours on any day.The manufacturer can sell a package of screw A at a profit of Rs7 and screw B at profit of Rs10.Assuming that he can sell all the screws he manufactures,how many packages of each type should the factory owner produce in a day in order to maximize his profit?Determine the maximum profit.

Updated On: Oct 5, 2023
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Solution and Explanation

Let the factory manufacture x screws of type A and y screws of type B on each day.

Therefore, x≥0,y≥0 The given information can be compiled in a table as follows.

 

Screw A

Screw B

Availability

Automatic machine(min)464×60=120
Hand Operated Machine(min)634×60=120

The profit on a package of screws A is Rs7 on the package of screws B is Rs10.

Therefore,the constraints are 4x+6y≤240 6x+3y≤240 Total profit,Z=7x+10y

The mathematical formulation of the given problem is
Maximize Z=7x+10y....(1)

Subject to the constraints,
4x+6y≤240...(2)
6x+3y≤240...(3)
x,y≥0...(4)

The feasible region determined by the system of constraints is The corner points are A(40,0),B(30,20),and C(0,40).

The values of Z at these corner points are as follows.

The feasible region determined by the system

The maximum value of Z is 410 at (30,20).

Thus, the factory should produce 30 packages of screw A and 20 packages of screw B to get a maximum profit of Rs410.

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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.