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Minimisation question - volume

  1. Feb 1, 2012 #1
    A coal box, in the shape of a cuboid, is to be placed flush against a wall so that only its top, front and two ends are visible. How should the height h and the depth d be chosen so a to minimise the visible surface area A under the constraint that the box must be able to contain atleast a certain volume V of coal?

    Here's how far I've got:

    V=hwd (where w is the width)
    Visible surafce area = 2hd + hw +wd
     
  2. jcsd
  3. Feb 1, 2012 #2

    HallsofIvy

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    There are two ways to approach this. One is to use hwd= V to reduce the number of variables: h= V/wd so the function to be minimized becomes 2hd + hw +wd = 2(V/wd)d+ (V/wd)w+ wd= 2V/w+ V/d+ wd. Now take the partial derivatives with respect to w and d and set them equal to 0.

    The other way is to use Lagrange multipliers. The gradient of the target function is <2h+ w, 2d+ w, h+ d> where the components are the derivatives with respect to d, h, and w in that order. The gradient of the constraint function, with derivatives in the same order, is <hw, wd, hd>. At the optimal point, we must have [itex]<2h+ w, 2d+ w, h+ d>= \lambda<hw, wd, hd>[/itex]. That is, we must have [itex]2h+ w= \lambda hw[/itex], [itex]2d+w= \lambda wd[/itex], and [itex]h+ d= \lambda hd[/itex], which, together with hwd= V, give four equations for d, h, w, and [itex]\lambda[/itex].

    Since a value of [itex]\lambda[/itex] is not necessary for the solution, I find it is often best to eliminate [itex]\lambda[/itex] by dividing one equation by another.
     
  4. Feb 1, 2012 #3
    Thanks - I'll give both methods a go and see how I get on!
     
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