Posted by **sh** on Wednesday, December 8, 2010 at 9:45pm.

To construct a tincan, V=32pi m^3, The cost per square meter of the side is half of the top and bottom of can. What are the dimensions and the cost?

V=πr²h=32pi SA=2πr²+2πrh

h=32/r²

Domain={r>o}

Let x be the cost, I subbed 32/r² for h

C=f(r)=2πxr²+(32xπ/r)

f'(r)=4πxr + 32xπ/r²

common denominator

f'(r)=(4πxr³+ 32xπ)/r²

I factored, it, r=-2, which is not in the domain.

Where did it go wrong?

- Calculus -
**Reiny**, Wednesday, December 8, 2010 at 10:58pm
Cost = 2(2πr^2) + 1(2πrh)

= 4πr^2 + 2πr(32/r^2)

= 4πr^2 + 64π/r

d(Cost)/dr = 8πr - 64π/r^2 = 0 for a min of Cost

8πr = 64π/r^2

r^3 = 8

r = 2

h = 32/4 = 8

Can should have a radius of 2 and a height of 8

- Calculus -
**sh**, Wednesday, December 8, 2010 at 11:13pm
I did the derivative wrong! Thanks!

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