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Multivariable calculus: Maximization of volume of box with constraint

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the maximum possible volume of a rectangular box inscribed in a hemisphere of radius R? Assume that one face of the box lies in the planar base of the hemisphere.

    NOTE: For this problem, we're not allowed to use Lagrange multipliers, since we technically haven't learned them yet. I know this is a pain, but please help me out!
    2. Relevant equations

    [tex] Volume\, of\, hemisphere\, = \frac{2}{3}R^{3}
    \\ Volume\, of\, rectangular\, box=xyz [/tex]
    (assuming width x, length y, height z)

    3. The attempt at a solution
    This is what I've written on my sheet so far:
    Assume x,yz is the intersection point of the top vertex of the box with the curved surface of the hemisphere in quadrant I.
    [tex] \Rightarrow\text{Dimensions of box}=2x\times2y\times z
    \\ \Rightarrow\text{Volume of box}=4xyz [/tex]
    If hemisphere has radius R,
    [tex] R=\sqrt{x^{2}+y^{2}+z^{2}}
    \\ \Rightarrow R^{2}=x^{2}+y^{2}+z^{2} [/tex]
    where R is constant for the purposes of maximization.

    Therefore, maximize [itex] f\left(x,y,z\right)=4xyz [/itex] with constraint [itex]R^{2}=x^{2}+y^{2}+z^{2}[/itex]

    [tex] f_{x}=4yz
    \\ f_{y}=4xz
    \\ f_{z}=4xy
    \\ f_{x}=f_{y}=f_{z}=0
    \\ \Rightarrow x=y=z=0 [/tex]
    Since this constitutes a non-existant box, the interior critical point is not the global maximum. We must find critical points on the boundary.

    Here is where I'm stuck :confused:. How do you find critical points on this kind of boundary function? Do you convert to polar coordinates? Obviously the coordinates are going to be in terms of R, but how do you eliminate x,y,z from the equations describing the coordinates?
     
    Last edited: Mar 27, 2012
  2. jcsd
  3. Mar 27, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why 4xyz? Since each edge has length twice the corresponding coordinate, shouldn't the volume be 8xyz?

    Also, what you have done completely ignores the constraint. Since, allowing x, y, and z to be any numbers, the box can have arbitrarily large volume what you found was the only extremum, the minimum possible volume.

    Since you are not allowed to use Lagrange multipliers, the best thing to do is to use the constraint to get, say, [itex]z= \sqrt{R^2- x^2- y^2}[/itex] and put that into the formula for volume of the box: [itex]V= 8xyz= 8xy(R^2- x^2- y^2)^{1/2}[/itex]. Differentiate that with respect to x and y and set the derivatives equal to 0.
     
  4. Mar 27, 2012 #3
    HallsofIvy, you're completely correct in that I ignored the constraint. My main problem was I had a mental gap in figuring out how to integrate the extremum into the equation. Thanks for the help though, now I've got my answer!

    Also, note that the question states that the box is contained within a "hemisphere". Since we are assuming that the box is centered at the origin, and the hemisphere is radiated outwards from the origin, we can safely assume that 2 dimensions will have twice the length of their radial components (going "horizontally across" the dome). However the third dimension of the box extends only to the base of the hemisphere (goes "vertically up" the dome). If the third dimension of the box was also twice the length of its radial component, the box would be contained within a sphere, not a hemisphere.

    I appreciate the response though! :smile:
     
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