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Best dimensions for maximum surface area

  1. Dec 27, 2006 #1
    I was doing some math problems involving surface area and maximum dimensions and then I wondered:
    Suppose you are given the surface area of a rectangular box but none of its dimensions. Is it possible to find the best dimensions (x,y,z) that would give the maximum volume of the box? I was thinking of something along the lines of making a graph and finding its maximum value ... Ex. 400cm² = 2xy + 2xz + 2yz. Possible? Unless there are other ways of doing it...
  2. jcsd
  3. Dec 28, 2006 #2


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    Make a graph of what? Not the surface area function that you give: you want to maximize the volume: V= xyz.

    The standard way of "maximizing" (or "minimizing") a function is to find its derivative and set that equal to 0. Since here you have a function of three variables (x,y,z) and an additional constraint, there are two ways to do it.

    1) Use the constraint to reduce the number of variables: 2xy+ 2xz+ 2yz= 400 so 2(x+y)z= 400- 2xy so z= (200-xy)/(x+y) and then V= xy(200-xy)/(x+y). Differentiate that with respect to x and y (partial derivatives) and set them equal to 0. You can probably see those are going to be messy derivatives!

    2) The Lagrange Multiplier method: At a maximum (or minimum) value, the gradient of the function must be a multiple of the gradient of the constraint. Here the function to be maximized is V= xyz. grad V= yzi+ xzj+ xyk. The constraint function is U= 2xy+ 2xz+ 2yz= 400 and grad U= (2y+ 2z)i+ (2x+2z)j+ (2x+2y)k. One is a multiple of the other if [itex]yz= \lambda(2y+ 2z)[/itex], [itex]xz= \lamba(2x+2z)[/itex], and [itex]xy= \lambda(2x+ 2y)[/itex]. That [itex]\lambda[/itex] is the "Lagrange Multiplier".
    If you eliminate [itex]\lambda[/itex] from those, you can get immediately that x= y= z: a cube will maximize volume for given surface area.
    Last edited: Dec 28, 2006
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