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
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?
2 answers
I did the derivative wrong! Thanks!
3 answers