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!

Maximum area for inscribed cylinder

  1. May 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Inscribe in a given cone, the height h of which is equal to the radius r of the base, a cylinder (c) whose total area is a maximum. Radius of cylinder is rc and height of cylinder is hc.

    2. Relevant equations
    A = 2πrchc + 2πrc2

    3. The attempt at a solution
    r = h ∴ hc = r - rc
    A = 2πrc(r - rc) + 2πrc2

    To get the maximum of this area I will find the radius rc when the growth of the area is zero.
    dA/drc = 0 = 2πr

    What does this result mean? I don't understand how 0 = 2πr makes sense as a result from a derivation. What kind of information does this result give me geometrically? How can I know that there is no maximum area to the cylinder as my answer sheet tells me.
  2. jcsd
  3. May 13, 2017 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Check your result for dA/drc

    [edit] oops, I checked my result and now we agree. Looks as if the area does not depend on rc -- as I could have read in your post :rolleyes:

    [edit] oops2 Thanks Ray for putting me right -- dA/drc does not depend on rc but the area itself of course does. It just keeps increasing until rc hits r and then you have a pancake, not a cylinder any more. Tricky exercise if you are doing it before coffee o0)
    Last edited: May 13, 2017
  4. May 13, 2017 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your problem is constrained: ##\max f = r_c h_c + r_c^2## subject to ##r_c+h_c = r##. The objective is ##A/(2 \pi)## and the constraint arises because the cylinder is inscribed in a cone of base-radius and height ##r##. You can substitute ##h_c = r - r_c## into ##f## to get a one-dimensional problem ##\max f_1 = r_c^2 + r_c(r-r_c) = r r_c,## subject to ##0 \leq r_c \leq r##. What is the solution to this last problem? (Remember that ##r## is a constant, so ##r_c## is the only variable.)
  5. May 13, 2017 #4
    Sorry, bad title. What I'm looking for is the maximum in the curve of the growth of area for a cylinder inscribed in a cone. Not the maximum area. What I'm hung up on the result of my derivation dA/drc = 0 = 2πr. If for example my cone had the radius r of 1 meter than my derivation would give dA/drc = 2πr = 6.28 = 0 which doesn't make sense. But dA/drc = 0 is just a statement I made and if dA/drc ≠ 0 then that means my statement is false and there is no maximum or minimum in the growth of area for the cylinder. Seems like I figured it out while writing me reply... Sorry for wasting your time and thank you all for trying to help me!
  6. May 13, 2017 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Setting the derivative to zero is a mistake: you have a constrained problem! That is, your originally-described problem is
    $$\begin{array}{l}\max A(r_c) = 2 \pi r_c(r-r_c) + 2 \pi r_c^2 = k r_c, \\
    \text{subject to} \;\;0 \leq r_c \leq r.
    Here, ##k = 2 \pi r## is a positive constant.

    Your newly-described problem is to maximize the constant ##k## on the set ##0 \leq r_c \leq r##, which makes no sense: the area just keeps growing at a steady rate, so every point is a maximum growth rate point.

    Setting a derivative to zero is what you do when you are looking for an interior-point max or min of a function, but not when you are on the boundary of an inequality constraint set. You would never, ever solve ##\max/ \min f(x) = 2x, \: 0 \leq x \leq 1## by setting the derivative of ##2x## to zero.
    Last edited: May 13, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted