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November 3rd, 2009, 07:17 AM   #1
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Simple Mathematical Finance problem - please help...

Seems simple but I'm stuck...

A company faces the following problem: the material for a rectangular container costs 3 per square foot for the bottom and 2 per square foot for the sides and top. Find the dimensions of the container so that its V=12 cubic feet and the cost is minimum.

Help much appreciated!
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November 3rd, 2009, 01:47 PM   #2
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Re: Simple Mathematical Finance problem - please help...

Quote:
Originally Posted by jamie89
Seems simple but I'm stuck...

A company faces the following problem: the material for a rectangular container costs 3 per square foot for the bottom and 2 per square foot for the sides and top. Find the dimensions of the container so that its V=12 cubic feet and the cost is minimum.

Help much appreciated!
Let x and y be the dimensions of the bottom, and z the height.

The cost of the top + bottom is 5xy, while the cost of the side is 4z(x+y). The condition is xyz=12. The problem then is to minimize:
5xy + 48(x+y)/(xy)=5xy + 48(1/x + 1/y)

Taking partials with respect to x and y and setting = 0 leads to:
5y = 48/(x^2) and 5x=48/(y^2)

I worked it out (no guaranties!!) and got x=y=(48/5)^1/3 and z=12(5/4^2/3
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