Optimization: 500 cu ft tank, square base, open top, welded

ptsoccerboy

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Aug 29, 2007
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I really need help on this problem.

"Your iron works has contracted to design and build a 500-ft^3, square-based, open top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible.

a. What dimensions do you tell the shop to use?
b. Briefly describe how you took weight into account.

1) So far, I set up an equation (x^2)(y)=500
2) Then I tried taking the derivative using the product rule and I got (dy/dx) = (-1000/(x^3))

I think that's the right derivative but I'm not sure where to go from there. I think I have to find critical points from the derivative (if that's the case, I know 0 is a critical point) but then if I evaluate my first equation for 0, I just get 0 and that gets me nowhere.

I checked the answer in the back of my textbook and it says the dimensions should be 10 ft by 10 ft by 5 ft but i dont know how they got it. Please help. Thank you!
 
Re: Optimization

Double check how you calculated the derivative. Remember, product rule is x'y+xy'... find y'. and the derivative of a constant is 0.
 
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