1. ## Linear Programming...

Certain animals in a rescue shelter must have at least 30g of protein and at least 20g of fat per feeding period. These nutrients come from food A, which costs 18 cents per unit and supplies 2g of protein and 4g of fat; and food B, which costs 12 cents per unit and has 6g of protein and 2g of fat. Food B is bought under a long–term contract requiring that at least 2 units of B be used per serving. How much of each food must be bought to produce the minimum cost
per serving?

2. That's a great problem. What are your plans for solving it?

Set up some coordinate axes.

Write the equations of your constraints.

Think about your minimization formula. Just pick a few values and graph the resulting lines. See if it soaks in.

3. Originally Posted by tuheetuhee
Certain animals in a rescue shelter must have at least 30g of protein and at least 20g of fat per feeding period. These nutrients come from food A, which costs 18 cents per unit and supplies 2g of protein and 4g of fat; and food B, which costs 12 cents per unit and has 6g of protein and 2g of fat. Food B is bought under a long–term contract requiring that at least 2 units of B be used per serving. How much of each food must be bought to produce the minimum cost
per serving?

I found this yesterday before my internet connection left me disconnected for eternity. I wanted to do this for practice yesterday. Let me do it now.

For protein, per serving:
at least 30 g.
A(2) +B(6) >= 30
A +3B >= 15 -----------(1)

For fat, per serving:
at least 20 g.
A(4) +B(2) >= 20
2A +B >= 10 -----------(2)

Food B is bought under a long–term contract requiring that at least 2 units of B be used per serving.
So, B >= 2 --------------------(3)

Plot those 3 inequalities on the same cartesian (A,B) axes.

Solve for their intersection points. These will be the vertices of the feasible region.

Between (1) and (2),
A +3B >= 15 -----------(1)
2A +B >= 10 -----------(2)

(1)*2, then minus (2),
5B = 20 -----so, B = 4, and then A = 3
Hence, intersection point is (3,4)

Between (2) and (3), intersection point is (4,2).
Between (1) and (3), intersection point is (9,2).

In the figure, the feasible region is an open one. It extends to infinity. It is bounded by the the 3 inequalities and there only two vertices: (3,4) and (9,2).

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How much of each food must be bought to produce the minimum cost
per serving?

per serving, Cost = A(0.18) +B(0.12) ---------(4)

Test that on the two vertices of the feasible region.
(In the graph, since the corner (3,4) is nearer to the origin (0,0) than the corner (9,2), then the minimum should be at the corner (3,4).)

At corner (3,4),
Cost = 3(0.18) +4(0.12) = 0.54 +0.48 = 1.02 dollars

At corner (9,2),
Cost = 9(0.18) +2(0.12) = 1.86 dollars.

$1.02 is less than$1.86.
Therefore, per serving, for minimum cost, buy 54 cents of food A and 48 cents of food B. ----answer.