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Spectre32
Mar18-04, 08:08 PM
Ok my teacher was going over this today, but i have no idea how to solve this.

A Closed box with a square base is to have a volume of 16000cm^3. The top and bottme cost 3 persq/cm. while the top is 1.50 pr sq/cm Find the demison of the box that will lead to the total minium total cost. WHat is the total cost.

cookiemonster
Mar18-04, 08:44 PM
So does the top cost 3/cm^2 or 1.5/cm^2?

cookiemonster

Spectre32
Mar18-04, 09:14 PM
Whoops. my bad. THe top and bottem are $3per sq/cm and the sides are $1.5 per sq/cm

cookiemonster
Mar18-04, 09:21 PM
Okay. The box is a rectangular prism. It has a base of dimensions length (l) and width (w). It also has height (h). Its volume is defined by

V = l*w*h

which is required to be 16000cm^3, so

lwh = 16000.

Additionally, the cost of the box is given by

C = $1.5*2(lh + wh) + $3*2(lw)

Using the equation for volume you can eliminate one of the variables, giving you the cost in terms of only 2 variables. You'll have to minimize this function using techniques you learned in class.

Why don't you give it a shot and post what you get?

cookiemonster

Chen
Mar19-04, 01:55 AM
I thought the box was a square prism ("A Closed box with a square base")...

cookiemonster
Mar19-04, 02:02 AM
So it is. 3 mistakes in 1 day. Go me.

It's easy to fix. Just let l = w and minimize the remaining variable.

cookiemonster

Chen
Mar19-04, 02:20 AM
Is there actually an answer to the question if the box was not a square prism, and no other data was supplied? Can't see how myself. [:)]

cookiemonster
Mar19-04, 02:48 AM
I don't think it matters either way. The top and bottom contribute most to the cost, so minimizing those areas will likely minimize the cost as well. It just makes the math a bit more difficult.

Then again, I haven't run the numbers, so I could be (and considering the day, probably am) wrong.

cookiemonster

matt grime
Mar19-04, 03:27 AM
You have to at least state what kind of aperture you want on the 'box'. The (smooth) surface with maximal volume per unit of surface area is the sphere; but where do you put stuff in? These problems are (often) solvable with calculus of variations (probably) subject to the smoothness constraints etc.