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(Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.


Solution

Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \geq 0,y \geq 0$$. We make the following table from the given data:
ResourcesFood
I
$$(x)$$
Food
II
$$(y)$$
Requirement
Vitamins A (units/ kg)$$2$$$$1$$$$8$$
Vitaminc C (units/ kg)$$1$$$$2$$$$10$$
Cost (Rs/ kg)$$50$$$$70$$
Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:
$$2x + y \geq 8$$
$$x + 2y \geq 10$$
Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is
$$Z = 50x 70y$$
Hence, the mathematical formulation of the problem is:
Minimise $$Z = 50x + 70y ... (1)$$
subject to the constraints :
$$2x + y \geq 8 ... (2)$$
$$x + 2y \geq 10 ... (3)$$
$$x, y \geq 0 ... (4)$$
Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.
Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.
Corner Point$$Z = 50x + 70y$$
$$(0, 8)$$$$560$$
$$(2, 4)$$$$380\leftarrow Minimum$$
$$(10, 0)$$$$500$$
In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality
$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$
to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.
Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.
802531_846816_ans_a6de210e20f74902a792b02a49fe6ca5.jpg

Mathematics

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