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The production manager for Beer etc. produces 2 kinds of beer: light (L) and dark (D). Two resources used to produce beer are malt and wheat. He can get at most 4800 oz of malt per week and 3200 oz of wheat per week. Each bottle of light beer requires 12 oz of malt and 4 oz of wheat, while a bottle of dark beer uses 8 oz of malt and 8 oz of wheat. Profits for light beer are $2 per bottle, and profits for dark beer are $1 per bottle.

2007-02-24 01:45:33 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

max 2L + D
where
L = 12M + 4W
D = 8M + 8 W
M < 4800
W < 3200

It's a linear programming optimization problem. You can probably find free LP software on the web

2007-02-24 02:11:22 · answer #1 · answered by Net Rider 3 · 0 1

I always mess these equations up (they seem like I'm writing them backwards). The following is the amount of malt and wheat as a function of the number of light and dark beers:
0 <= M = 12L + 8D <= 4800
0 <= W = 4L + 8D <= 3200

P = 2L + 1D (the objective function)

Graphing the two functions (12x + 8y < 4800 and 4x + 8y < 3200) and using the x-axis and y-axis as additional boundaries, you'll see our quadrilateral region to investigate. We only need to check the corners, so that would be at:
(0,400), (400,0), (0,0), and (200,300)

Checking with the objective function:
2*0+1*400 = 400
2*400+1*0 = 800
2*0 + 1*0 = 0
2*200 + 1*300 = 700

So the maximum profit is obtained when 400 Light bears are sold and 0 Dark bears are sold.

2007-02-24 02:30:28 · answer #2 · answered by Anonymous · 0 1

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