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?
Let x be the cost, I subbed 32/r² for h
f'(r)=4πxr + 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!
Answer This Question
More Related Questions
- Calculus - an open topped cylinder has a volume of 125 cubic inches. determine ...
- Calculus - How do I find the critical values? y= 4/x + tan(πx/8) What I ...
- Math, please help - Which of the following are trigonometric identities? (Can be...
- geometry/algebra - I asked this question yesterday, and I'm trying to understand...
- Math integrals - What is the indefinite integral of ∫ [sin (π/x)]/ x^...
- Trig - find all solutions to the equation √3 csc(2theta)=-2 Would the ...
- math plz help ;-; - Assuming that the cone is completely filled with ice cream, ...
- Trig Help! - Question: Trying to find cos π/12, if cos π/6 = square ...
- calculus - A manufacturer wishes to produce the most economical packaging for ...
- PreCalc - Graph y=2csc(x-π) New period that I solve for 0<x-π<2...