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Math Help - 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:

    V = \pi r^2h

    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:

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


    From equation I, h = 2r.

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

    Now I am stuck here.
    Last edited by Keep; May 20th 2010 at 03:15 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 " 2rh= 4r" and " 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 r^2= 2r immediately implies r= 0 or r= 2 (which is not a solution)? And your third "equation", 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 \nabla F= \lambda\nabla G for some constant \lambda.

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

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

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

    \nabla F= \lambda \nabla G is, then,
    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
    2\pi rh= 2\lambda\pi (r+ h) and \pi r^2= 2\pi\lambda r.

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

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

    Solve that, together with \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:

    V =\pi r^2 h
    A = \pi r^2 + \pi rh
    <br />
C = \pi r^2 + \pi rh

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

    <br />
2\pi rh = 2\pi r\lambda
    <br />
\pi r^2 = \pi r\lambda
    <br />
C = \pi r^2 + \pi rh


    dividing equation I by 2 we have:

    <br />
\pi rh = \pi r\lambda
    \pi r^2 = \pi r\lambda
    C = \pi r^2 + \pi rh

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

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

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

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


    Am I on the right track so far?
    Last edited by Keep; May 21st 2010 at 03:29 PM.
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