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Question
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:
| Vitamin A |
Vitamin B |
Vitamin |
Vitamin D |
|
| Food X Food Y |
1 2 |
1 1 |
1 3 |
2 1 |
One kg food X costs Rs 5, whereas one kg of food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.
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Solution
Let the dietician wishes to mix x kg of food X and y kg of food Y.
Therefore,
As we are given,
| Vitamin A | Vitamin B |
Vitamin C |
Vitamin D | |
| Food X Food Y |
1 2 |
1 1 |
1 3 |
2 1 |
It is given that the mixture should contain at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D.
Therefore, the constraints are
\[x + 2y \geq 6\]
\[x + y \geq 7\]
\[x + 3y \geq 11\]
\[2x + y \geq 9\]
It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs 8 per kg.
Thus, Z = \[5x + 8y\]
Thus, the mathematical formulation of the given linear programmimg problem is
Minimize Z = \[5x + 8y\]
subject to
\[x + 2y \geq 6\]
\[x + y \geq 7\]
\[x + 3y \geq 11\]
\[2x + y \geq 9\]
First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 6, x + y = 7, x + 3y =11, 2x + y =9, x = 0 and y = 0.
The line x + 2y = 6 meets the coordinate axis at A1(6, 0) and B1(0, 3). Join these points to obtain the line x + 2y = 6. Clearly, (0, 0) does not satisfies the inequation x + 2y ≥ 6. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line x + y = 7 meets the coordinate axis at C1(7, 0) and D1(0, 7). Join these points to obtain the line x + y = 7. Clearly, (0, 0) does not satisfies the inequation x + y ≥ 7. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line x + 3y = 11 meets the coordinate axis at \[E_1 \left( 11, 0 \right)\] and \[F_1 \left( 0, \frac{11}{3} \right)\] Join these points to obtain the line x + 3y = 11.Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 11. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line 2x + y = 9 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 H1(0, 9), I1(2 ,5), J1(5, 2), E1(11, 0).
The values of Z at these corner points are as follows
| Corner point | Z= 5x + 8y |
| H1 | 72 |
| I1 | 50 |
| J1 | 41 |
| E1 | 55 |
The minimum value of Z is at J1(5, 2) which is Rs 41.
Hence, cheapest combination of foods will be 5 units of food X and 2 units of food Y.
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