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Thread: Cylinder with largest volume

  1. #1
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    Cylinder with largest volume

    Given cylinders of equal surface area, fnd the one with largest Volume.

    Attempt:

    $\displaystyle V = \pi r^2h$

    $\displaystyle A = 2\pi r^2 + 2 \pi rh$

    Now I want to use Lagrange multiplier and use the area as a constraint. However, the pi is scaring me. Pi is a constant and so far I have only learnt doing it with variables. I thought of substituting pi with x for example but that is variable. Or should I just treat it as constant and go on with it like this:

    $\displaystyle 2rh = 4r$ .......................... I
    $\displaystyle r^2 = 2r$ ........................... II
    $\displaystyle A = 2\pi r^2 + 2 \pi rh$ ........................... III


    From equation I, h = 2r.

    Substituting in III, we get $\displaystyle 4\pi r^3 + 4\pi r^2$, which is the same as $\displaystyle \pi r^3 + \pi r^2$

    Now I am stuck here.
    Last edited by Keep; May 20th 2010 at 02:15 AM.
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  2. #2
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    This is very strange. You are doing Calculus yet you say "However, the pi is scaring me. Pi is a constant and so far I have only learnt doing it with variables." Have none of the problems you have done involve "2" or "3"? Those are also constants. Even if you had a problem with, say , "xyz", there is a "1" implied that is a constant.

    You then have "$\displaystyle 2rh= 4r$" and "$\displaystyle r^2= 2r$". Where did those come from? You say you are using the "Lagrange multiplier method" but if so where is the multiplier? Yes, 2rh= 4r implies h= 2r but did you notice that $\displaystyle r^2= 2r$ immediately implies r= 0 or r= 2 (which is not a solution)? And your third "equation", $\displaystyle 2\pi r^2h+ 2\pi r^2+ 2\pi rh$ is not even an equation!

    The idea of the "lagrange Multiplier Method" is that any maximum or minimum of F(x,y,z) subject to the constraint G(x,y,z)= constant occurs where $\displaystyle \nabla F= \lambda\nabla G$ for some constant $\displaystyle \lambda$.

    Here, F is $\displaystyle \pi r^2h$, the volume function, and G is $\displaystyle \pi r^2+ 2\pi rh$, the total surface area.

    $\displaystyle \nabla F= 2\pi rh\vec{i}+ \pi r^2\vec{j}$, where I have arbitrarily written $\displaystyle \vec{i}$ for a unit vector in the "r direction" and $\displaystyle \vec{j}$ for a unit vector in the "h direction".

    $\displaystyle \nabla G= (2\pi r+ 2\pi h)\vec{i}+ 2\pi r\vec{h}$.

    $\displaystyle \nabla F= \lambda \nabla G$ is, then,
    $\displaystyle 2\pi rh\vec{i}+ \pi r^2\vec{j}= (2\pi r+ 2\pi h)\vec{i}+ 2\pi r\vec{h}$

    Setting the components equal we have the two equations
    $\displaystyle 2\pi rh= 2\lambda\pi (r+ h)$ and $\displaystyle \pi r^2= 2\pi\lambda r$.

    Those, together with the equation $\displaystyle \pi r^2+ 2\pi rh= C$, give three equations to solve for r, h, and $\displaystyle \lambda$.

    Since it is not necessary to find $\displaystyle \lambda$ to solve this problem, it is often simplest to eliminate $\displaystyle \lambda$ by dividing one equation by another:
    $\displaystyle \frac{2\pi rh}{\pi r^2}= \frac{2\pi (r+h)}{2\pi r}$
    which reduces to
    $\displaystyle \frac{2h}{r}= \frac{r+h}{r}$.

    Solve that, together with $\displaystyle \pi r^2+ 2\pi rh= C$.
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  3. #3
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    Thanks. Yeah, you are right, it is really strange. I should have seen pi just like any other constant. Anyway, here is what I should have done:

    $\displaystyle V =\pi r^2 h $
    $\displaystyle A = \pi r^2 + \pi rh$
    $\displaystyle
    C = \pi r^2 + \pi rh$

    Then using Lagrange multiplier (which as you pointed out wasn't even there in my first post!)

    $\displaystyle
    2\pi rh = 2\pi r\lambda$
    $\displaystyle
    \pi r^2 = \pi r\lambda$
    $\displaystyle
    C = \pi r^2 + \pi rh$


    dividing equation I by 2 we have:

    $\displaystyle
    \pi rh = \pi r\lambda$
    $\displaystyle \pi r^2 = \pi r\lambda$
    $\displaystyle C = \pi r^2 + \pi rh$

    now we divide equation II by equation I we get $\displaystyle r^2 = rh$ which means $\displaystyle r = +- \sqrt {rh}$ which can also be got from your equation $\displaystyle \frac{2h}{r}= \frac{r+h}{r}$.

    Substituting for $\displaystyle r^2 = rh$ in III we get $\displaystyle C= 2\pi rh$ Now substitung for $\displaystyle r = \sqrt {rh}$ we get:

    $\displaystyle 2\pi rh = \pi rh + h\pi \sqrt {rh} $ which is the same as
    $\displaystyle rh = h\sqrt {rh} $

    this means $\displaystyle h = \frac{rh} {\sqrt {rh}} $


    Am I on the right track so far?
    Last edited by Keep; May 21st 2010 at 02:29 PM.
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