The length of a cedar chest is twice its width. The cost/dm^2 of the lid is four times the cost/dm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440dm^3, find the dimensions so that the cost is a minimum.

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- September 10th 2010, 10:09 AMeuclid2optimization problem2
The length of a cedar chest is twice its width. The cost/dm^2 of the lid is four times the cost/dm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440dm^3, find the dimensions so that the cost is a minimum.

- September 10th 2010, 10:23 AMundefined
- September 10th 2010, 10:33 AMeuclid2
i'm no longer posting my work. i posted all my work for my other problem and no help was provided. why waste my time?

- September 10th 2010, 10:55 AMundefined
1) You don't have to provide full work, you can describe and/or give a sketch

2) Sorry to hear that

3) The forum still has rules, even if for some questions help was never received

4) Mind providing the link for that other problem?

5) Try seeing from my perspective. A person doesn't put much effort into posting a question. Why should I put much effort into posting an answer?

6) My approach to the problem would be to write several equations

L = 2w

Lwh = 1440

C = 4Lw + 2hw + 2hL + Lw

Use the first two equations to substitute into the third so that you have C as a function of just one variable, take the derivative, set to 0, find the other two variables from the first. - September 10th 2010, 11:23 AMeuclid2
thank you sir. your response was very helpful.

link to the other problem: http://www.mathhelpforum.com/math-he...st-155752.html