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Diet Problems

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Diet Problems - Lesson Summary

Manufacturing problems:

Determine the number of units of different products that should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output, etc., in order to make maximum profit.

Diet problems:

In these problems, the amount of different kinds of nutrients and other constituents that should be included in a diet, so that the cost of the diet is minimum and it contains a certain minimum amount of each constituent/nutrients, is determined.

Ex: A housewife wishes to mix together two kinds of food, P and Q, 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 number of units of vitamin in one kg of each kind of food is given below:

  Vitamin A

  Vitamin B

  Vitamin C

  Food P

          1

          2

        3

  Food Q

          2

          2

        1

One kg of food P costs Rs. 6 and one kg of food Q costs Rs. 10. Find the minimum cost of the mixture that will produce the required diet.

Solution:

Let x kg of food P and y kg of food Q be mixed together.

⇒ x ≥ 0, y ≥ 0 … (1)

Food P has 1 unit of vitamin A, while food Q has 2 units of it.

Total vitamin A content = x + 2y

⇒ x + 2y ³ 10 ... (2)

Food P has 2 units of vitamin B and food Q also has 2 units of it.

Total vitamin B content = 2x + 2y

⇒ 2x + 2y ³ 12

⇒ x + y ³ 6 ... (3)

Similarly, food P has 3 units of vitamin C, while food Q has 1 unit of it.

Total vitamin C content = 3x + y

⇒ 3x + y ³ 8 ... (4)

One kg of food P costs Rs. 6 and one kg of food Q costs Rs. 10.

Cost of x kg of food P = Rs. 6x

Cost of y kg of food Q = Rs. 10y

Total cost of food P and Q = 6x + 10y

⇒ Z = 6x + 10y is an objective function

Z = 6x + 10y is minimised using the constraints

x ≥ 0, y ≥ 0 … (1)

x + 2y ³ 10 ... (2)

x + y ³ 6 ... (3)

3x + y ³ 8 ... (4)

x ≥ 0, y ≥ 0

x + 2y ³ 10

i) x + 2y = 10

     x

     0

    10

     y

     5

     0

The ordered pairs are (0, 5), (10, 0).

x + y ³ 6

ii) x + y = 6

    x

    0

    6

    y

    6

    0

The ordered pairs are (0, 6), (6, 0).

3x + y ³ 8

iii) 3x + y = 8

      x

      2

     1

      y

      2

     5

The ordered pairs are (2, 2), (1, 5)

The shaded region is the feasible region, which is open (unbounded).

The corner points of the feasible region are

A (10, 0), B (2, 4), C (1, 5) and D (0, 8).

    Corner points (x, y)

   Value of the objective function, Z = 6x + 10y

             A (10, 0)

  Z = 6 (10) + 10 (0) = 60

             B (2, 4)

  Z = 6 (2) + 10 (4) = 52

             C (1, 5)

  Z = 6 (1) + 10 (5) = 56

            D (0, 8)

  Z = 6 (0) + 10 (8) = 80







The minimum value of Z is 52, which occurs at the point, (2, 4).

As the feasible region is unbounded, we need to draw the graph of the inequality.

⇒6x + 10y < 52

⇒ 3x + 5y < 26

There is no common point between the feasible region and the open half plane of the inequality 3x + 5y < 26.

Clearly, the minimum value of Z is 52, attained at the point, (2, 4).

The minimum cost of the diet is Rs. 52 when 2 kilograms of food P and 4 kilograms of food Q are mixed.

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