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August 28, 2014

August 28, 2014

Posted by **Jacob** on Wednesday, March 13, 2013 at 5:04pm.

- Applied Calculus -
**Steve**, Wednesday, March 13, 2013 at 5:10pmwe've been through several of these box problems now.

Have you some ideas of how to proceed?

- Applied Calculus -
**Jacob**, Wednesday, March 13, 2013 at 5:12pmI had to miss my last week of Calculus due to some personal things, so I had not been through any lectures of optimization. It hurt me greatly, now I have a few of these that I don't know how to set up at all.

- Applied Calculus -
**Steve**, Wednesday, March 13, 2013 at 5:21pmbase is side x, height is h, so

hx^2 = 50

h = 50/x^2

cost of the box is

cost of sides + cost of bottom + cost of top

that should get you started.

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