Question 108466: A school dietitian wants to prepare a meal of meat and vegetables that has the lowest possible fat and that meets the FDA recommended daily allowance of iron and protein. The RDA minimums are 20 mg of iron and 45 g of protein. Each 3 oz serving of meat contains 45 g of protein, 10 mg of iron, and 4 g of fat. Each 1 cup serving of vegetables contains 9 g of protein, 6 mg of iron, and 2 g of fat.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A school dietitian wants to prepare a meal of meat and vegetables that has the lowest possible fat and that meets the FDA recommended daily allowance of iron and protein. The RDA minimums are 20 mg of iron and 45 g of protein. Each 3 oz serving of meat contains 45 g of protein, 10 mg of iron, and 4 g of fat. Each 1 cup serving of vegetables contains 9 g of protein, 6 mg of iron, and 2 g of fat.
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Let "m" be the meat service; Let "v" be the vegetable service
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INEQUALITIES:
Protein: 45m + 9v >=45
Iron: 10m + yv >=20
Fat: 4m+2v (this is the object function you are trying to minimize).
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Graph the following in the 1st quadrant:
v>= -5m+5
v>= (-5/3)m+(10/3)
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Find the point of intersection of the two boundary lines:
-5m+5=(-5/3)m+(10/3)
-15m+15 = -5m+10
10m = 5
m = (1/2)
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Substitute to solve for v:
v = -5(1/2)+5
v = 5/2
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Substitute those values into the object function
Fat = 4(1/2) + 2(5/2) = 7 grams
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She should serve (1/2)(3oz) meat portion and a (5/2)(cup) vegetable portion.
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Cheers,
Stan H.
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