Linear Programming, Algebra 2 Question

Question

In order to ensure optimal health and accurate test results, a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than 5 ounces of food a day. Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbs, and 2 g of protein per ounce, and costs $.20/oz. Food Y contains 12 g of fat, 12 g of carbs, and 1 g of protein per ounce, and costs $.30/oz. What is the optimal blend to minimize cost?

Answer 1

min .2X + .3Y (Cost) subject to

8X + 12Y >= 24 (Fat)

12X + 12Y >= 36 (Carbs)

2X + 1Y >= 4 (Protein)

X + Y <= 5 (the rabbit can eat at most 5 ounces, so the inequality is in the opposite direction)

You can solve this graphically or with an LP solver. However, note that you can also solve this with a little thought.

If 12X + 12Y >= 36, 8X + 8Y >= 24, so 8X + 12Y >= 24; thus, 8X+12Y is dominated.

If 12X + 12Y >= 36, X + Y>= 3

If X + Y >= 3 and 2X + Y >=4, you would never reach 5 ounces, as you would never have X + Y > 4.

Then, if we seek to minimize .2X + .3Y with X + Y >= 3 and 2X + Y >= 4, if we just look at minimize .2X + .3y with X + Y >= 3, the more X the better, so let X = 3 and Y = 0. Then, we have satisfied 2x + Y >= 4

Thus, the solution is X = 3oz, Y = 0 oz.