Minimisation question - volume

[nathangrand]

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 + homework +wd

 

[HallsofIvy]

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 + homework +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 &lt;2h+ w, 2d+ w, h+ d&gt;= \lambda&lt;hw, wd, hd&gt;. That is, we must have 2h+ w= \lambda hw, 2d+w= \lambda wd, and h+ d= \lambda hd, which, together with hwd= V, give four equations for d, h, w, and \lambda.

Since a value of \lambda is not necessary for the solution, I find it is often best to eliminate \lambda by dividing one equation by another.

 

[nathangrand]

Thanks - I'll give both methods a go and see how I get on!