A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs250 per bag contains 3 units of nutritional element A,2.5units of element B and 2units of element C.Brand Q costing Rs200 per bag contains 1.5units of element B and 3units of element C. The minimum requirements of nutrients A,B, and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand that should be mixed in order to produce a mixture having a minimum cost per bag. What is the minimum cost of the mixture per bag?
Solution and Explanation
Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows.
The given problem can be formulated as follows.
Minimize Z=250x+200y.....(1)
Subject to the constraints,
3x+1.5y≥18.....(2)
2.5x+11.25y≥45....(3)
2x+3y≥24....(4)
x,y≥0.....(5)
The feasible region determined by the system of constraints is as follows.
The corner points of the feasible region are A(18,0), B(9,2), C(3,6) and D(0,12).
The value of Z at these corner points is as follows.
As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of Z.
For this, we draw a graph for inequality 250x+200y<1950 or 5x+4y<39, and check whether the resulting half-plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x+4y<39
Therefore, the minimum value of Z is 2000 at (3,6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950.
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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.