optimization: minimize material costs of cylindrical can

[xc630]

Hi, I would appreciate some help with this problem.

A cylindrical can must have a volume of 32pi cubic inches. What should be its radius and height in order to minimize the cost of the material (ignore leftovers) if the circular top and bottom each cost two cents per square inch and the lateral surface costs one cent per square inch?

So I know V= (pi)r^2*h and Lateral area = 2(pi)r*h

I solved V for h and plugged in LA getting 64(pi)/r. However when I tried solving its derivative for zero I couldn't. I also tried solving the derivative of 2(pi)r^2 (the tops?) + 64(pi)/r for zero and got r+ the cubic root of 16. The answer however should be r= 2 according to the answer key. Some help please!

 

[soroban]

Re: optimization

Hello, xc630!

A cylindrical can must have a volume of \(\displaystyle 32\pi\text{ in}^3.\)
What should be its radius and height in order to minimize the cost of the material
if the circular top and bottom each cost two cents per square inch
and the lateral surface costs one cent per square inch?

\(\displaystyle \text{The volume of the cylinder is: }\;V \;=\;\pi r^2h \:=\:32\pi\quad\Rightarrow\quad h \:=\:\frac{32}{r^2}\) .[1]

The top and bottom have an area of: \(\displaystyle 2\pi r^2\)
. . At 2 cents per square inch, its cost is: \(\displaystyle 4\pi r^2\) cents.

The lateral surface is: \(\displaystyle 2\pi rh\)
. . At 1 cent per square inch, its cost is: \(\displaystyle 2\pi rh\) cents.

\(\displaystyle \text{The total cost is: }\;C \;=\;4\pi r^2 + 2\pi rh\) . [2]


Substitute [1] into [2]:

. . . \(\displaystyle C \;=\;4\pi r^2 + 2\pi r\left(\frac{32}{r^2}\right) \;=\;4\pi r^2 + 64\pi r^{-1}\)


Now try it . . .

 

[xc630]

Re: optimization

thanks soroban