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Question
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements
Solution
Let the dietician wishes to mix x units of food F1 and y units of food F2.
Clearly, \[x, y \geq 0\]
The given information can be tabulated as follows:
| Vitamin A | Minerals | |
| Food F1 | 3 | 4 |
| Food F2 | 6 | 3 |
| Minimum requirement | 80 | 100 |
The constraints are
\[3x + 6y \geq 80\]
\[4x + 3y \geq 100\]
It is given that cost of food F1 and F2 is Rs 4 and Rs 6 per unit respectively. Therefore, cost of x units of food F1 and y units of food F2 is Rs 4x and Rs 6y respectively.
Let Z denote the total cost
∴ Z = 4x + 6y
Thus, the mathematical formulation of the given linear programmimg problem is
Minimize
\[Z = 4x + 6y\] subject to
\[3x + 6y \geq 80\]
\[4x + 3y \geq 100\]
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 6y = 80, 4x + 3y = 100, x = 0 and y = 0
The line 3x + 6y = 80 meets the coordinate axis at
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The feasible region determined by the system of constraints is
The corner points are D
|
Corner point |
Z= 4x + 6y |
| D \[\left( 0, \frac{100}{3} \right)\] | 200 |
| E \[\left( 24, \frac{4}{3} \right)\] | 104 |
|
\[A\left( \frac{80}{3}, 0 \right)\]
|
\[\frac{320}{3}\]
|
The minimum value of Z is Rs 104 which is at E \[\left( 24, \frac{4}{3} \right)\]
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