Question

A nutritionist at the Medical Center has been asked to prepare a special diet for

certain patients. She has decided that the meals should be prepared from Foods A and B and that they

should contain a minimum of 400 mg of calcium, 10 mg of iron, and 40 mg of vitamin C. Each ounce of Food

A contains 30 mg of calcium, 1 mg of iron, 2 mg of vitamin C, and 3 mg of cholesterol. Each ounce of Food

B contains 25 mg of calcium, 0.5 mg of iron, 5 mg of vitamin C, and 5 mg of cholesterol.

(1) How many ounces of each type of food should be used in a meal so that the cholesterol content is

minimized and the minimum requirements of calcium, iron, and vitamin C are met? What is the

minimum cholesterol content?

(2) By how much can the cholesterol content of Food A vary before the optimal solution changes?

(3) By how much can the cholesterol content of Food B vary before the optimal solution changes?

Answer 1

1) Let, the type of food A is x and type of food B is y

Amount of calcium in food A is 30mg and in food B is 25mg

Minimum amount of calcium is 400 mg implies

Amount of iron in food A is 1mg and in food B is 0.5mg

Minimum amount of iron is 10 mg implies

Amount of vitamin C in food A is 2mg and in food B is 5mg

Minimum amount of vitamin C is 40 mg implies

Also,

The amount of cholestrol is 3mg in food A and 5mg in food B

Plot the above inequalities and find the feasible region

The end points of feasible region are (20,0), (10,4), (5,10), (0,20). Calculate the function value and select the minimum value from it

Hence, minimum cholestrol is achieved at the end point (10,4). Therefore, the minimum is achieved if 10 ounces of Food A and 4 Ounces of Food B

The minimum value of cholestrol is 50mg

2)

currently we have our minimizing function as C=3x+5y we want to know a range of coefficient of x such that our solution doesn't changes.

C=3x+5y => y=-3/5 x+ C/5 i.e. -3/5 and passing through those critical point, we know that (10,4) gives us the optimum solution.

so if C=kx+5y then slope of this line is -k/5 which should should have slope less than 30x+25y i.e. -30/25 and greater than

2x+5y=40 i.e. -2/5

i.e.

multiply by -5 and reverse inequality

so k can take any value from 2 to 6

basically you have to rotate the line passing from 10,4 until it doesn't hit the pheasible reason of inequality.

similarly for 3rd part

C=3x+ky has slope -3/k

mulitply by -5

take inverse

7.5

so K can take value between 2.5 and 7.5 for no change in solution