1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivative problem

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A company wants to build a cylindrical can with a total surface area of 100 square inches, which includes the bottom and top of the can. Find the dimensions that maximize the volume.

    2. Relevant equations

    volume of a cylinder = 2*pi*r^2*h
    surface area of cylinder = 2*pi*r^2 + 2*pi*r*h

    3. The attempt at a solution

    I wrote a huge wall of text detailing how I solved the problem, but when I went to post I had gotten logged out due to how much time I spent on the post. So I'm going to make this short and just explicate my process:

    I took the formula for constant surface area 100 inches squared and solved for the height in terms of the radius. Then, I plugged that equation (h in terms of r) into the volume and took the derivative of the volume with respect to the radius.

    I set the derivative equal to zero to find the radius that gives the max volume, and I got
    r = sqrt(50/(pi*3))

    Then I plugged r back into the equation for h in terms of r, and simplifying, I got:
    h = sqrt(150/pi) - sqrt(50/(3*pi))


    The answer sheet I have says that I should have gotten H = 2R. Indeed my answer reflects this, but should I have arrived at a general solution H = 2R instead of the exact answers I got? How would I go about solving this generally?

    Was my approach fine? Could someone maybe solve it for me in an alternate manner? I would really appreciate iany advice. Thank you.
  2. jcsd
  3. Apr 9, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    There a setting that stops you from being logged out like that ... though I used to write out my posts on a text editor and copy-and-paste it over ;)
    The question asks for the specific solution - the general relation would maximize the volume for any given surface area.
  4. Apr 9, 2013 #3


    User Avatar
    Science Advisor

    If surface area [itex]S= 2\pi R H+ 2\pi R^2 [/itex] and you want to maximize the volume [itex]V= \pi R^2H[/itex] then [itex]\nabla V= 2\pi RH\vec{i}+ \pi R^2\vec{j}[/itex] must be a multiple of [itex]\nabla S= (2\pi H+ 4\pi R)\vec{i}+ 2\pi R\vec{j}[/itex]. That means we must have [itex]2\pi RH= \lambda(2\pi H+ 4\pi R)[/itex] and [itex]\pi R^2= 2\lambda\pi R[/itex] for some constant [itex]\lambda[/itex]. We can eliminate [itex]\lambda[/itex] by dividing the first equation by the second:
    [tex]\frac{2\pi RH}{\pi R^2}= \frac{2\pi H+ 4\pi R}{2\pi R}[/tex]
    [tex]\frac{2H}{R}= \frac{H+ 2R}{R}[/tex]
    [tex]2H= H+ 2R[/tex]
    [tex]H= 2R[/tex]
  5. Apr 9, 2013 #4
    Hey thanks!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted