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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
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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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