A dietician wishes to mix together two. kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

$figure$
One kg of food X costs Rs.16 and one kg of food Y costs Rs. 20. Find the least cost of the mixture which will produce the required diet ?

Question

A dietician wishes to mix together two. kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

$figure$

One kg of food X costs Rs.16 and one kg of food Y costs Rs. 20. Find the least cost of the mixture which will produce the required diet ?

Accepted Answer

.: Let 'x' kg of food A and ‘y’ kg of food Y be mixed. We have the data given in the table.

$figure$

Total cost of the mixture is $Z = 16x + 20y$.

Thus, mathematical formulation of the problem is as follows:

Minimize : $Z = 16x + 20y$ ... (1)

Subject to constraints

$x + 2y \ge 10$ ...(2)

$2x + 2y \ge 12 \Leftrightarrow x + y \ge 6$ ...(3)

$3x + y \ge 8$ ...(4)

$x,y \ge 0$ ...(5)

${l_1}:x + 2y = 10;$ ${l_2}:x + y = 6;$

${l_3}:3x + y = 8$

Let us graph the inequalities (2) to (5).

The feasible region (shaded) determined by

$figure$

the constraints (2) to (5) is shown in the graph and is unbounded.

Let us evaluate Z at the comer points A(10, 0), P(2, 4), Q(1, 5),F(0, 8).

In the table, we find that the smallest value of Z is 112 at the point (2, 4).

Since the feasible region is unbounded,

so we have to draw the graph of the inequality
$16x + 20y < 112$ i.e.

, $4x + 5y < 28$

to check whether the resulting open half plane has any point common with the feasible region.

From graph, we see that it has no points in common.

Hence; the minimum cost = Rs. 112 when 2 kg of food X and 4 kg of food Y are mixed.

Linear Programming — Class 12 Maths Solution