A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs4 per unit food and F2 costs Rs6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs4 per unit food and F2 costs Rs6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.
Solution and Explanation
Let the diet contain
x units of food F1 and y units of food F2.
Therefore, x≥0 and y≥0
The given information can be compiled in a table as follows.
| Vitamin A(units) | Mineral(units) | Cost per unit (Rs) | |
| Food F1(x) | 3 | 4 | 4 |
| Food F2(y) | 6 | 3 | 6 |
| Requirement | 80 | 100 |
Therefore, the constraints are 3x+6y≥80 4x+3y≥100 x,y≥0
The feasible region determined by the constraints is as follows.
It can be seen that the feasible region is unbounded. The corner points of the feasible region are A(\(\frac{8}{3}\),0), B(2,\(\frac{1}{2}\)) and C(0,\(\frac{11}{2}\))
The corner points are A(\(\frac{80}{3}\),0), B(24,\(\frac{4}{3}\)) and C(0,\(\frac{100}{3}\)).
The values of Z at these corner points are as follows.
As the feasible region is unbounded,
therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph for the inequality, 4x+6y<104 or 2x+3y<52, 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 2x+3y<52
Therefore, the minimum cost of the mixture will be Rs 104.
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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.