1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook