A veterinarian has been asked to prepare a diet for a group of dogs to be used in a nutrition study at the School of Animal Science. It has been stipulated that each serving should be no larger than 10 oz and must contain at least 29 units of Nutrient I and 20 units of Nutrient II. The vet has decided that the diet may be prepared from two brands of dog food: Brand A and Brand B. Each ounce of Brand A contains 3 units of Nutrient I and 4 units of Nutrient II. Each ounce of Brand B contains 5 units of Nutrient I and 2 units of Nutrient II. Brand A costs 5 cents/oz and Brand B costs 6 cents/oz. Determine how many ounces of each brand of dog food should be used per serving to meet the given requirements at a minimum cost.

Costs are reduced when you have 3 ounces of brand A and 4 ounces of brand B for a cost of 39 cents / serving.

Explanation:

Yes:

x = ounces of brand A

y = ounces of brand B

The objective function and restrictions will be as follows:

Minimize cost function, C = 5x + 6y

This function is subject to:

x + y <= 10

3x + 5y> = 29 eq. 1

4x + 2y> = 20 eq1.2

x, y> = 0

We need to graph the constraints. First, x, y> = 0 their axes meet at x and y as limits. x + y <= 10 is a line from (0.10) to (10.0) and is equal to the area below the graph. The equation 3x + 5y> = 29 passes through (0.5.8), (29 / 3.0) and is equal to the area on the line. 4x + 2y> = 20 passes from (0,10), (0,5) and is equal to the area on the line.

The limits are delimited by (0.10), (10.0), (29 / 3.0). Solving the equations:

12x + 20y = 116 (multiplying eq. 1 by 4)

-12x - 6y = -60 (multiplying eq. 2 by -3)

14 y = 56

y = 4

4x + 8 = 20

x = 3

Replacing in the objective equation:

C = 5 (0) + 6 (10) = 60

C = 5 (29/3) + 6 (0) = 48.33

C = 5 (10) + 6 (0) = 50

C = 5 (3) + 6 (4) = 39

Costs are reduced when you have 3 ounces of brand A and 4 ounces of brand B for a cost of 39 cents / serving.

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