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

    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 -

    I did the derivative wrong! Thanks!

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