# Find the minimum & maximum area of a box!

• Apr 13th 2011, 08:53 AM
nukenuts
Find the minimum & maximum area of a box!
Here's another question Ive come across during revision

http://i55.tinypic.com/2poxs36.png

So

Volume is
$w*d*h = 32000$
$w^2*h = 32000$

$h = \frac{32000}{w^2}$

Total area is base of box + 4 sides -
$w^2 + 4wh$
$w^2 + \frac{128000}{w}$

And then I got the derivative of $f(x) = w^2 + \frac{128000}{w}$

And where this derivative = 0, there was a local minimum. This was where w = 40, so that makes d = 40 and h = 20 for the dimensions of the box that will use the minimum amount of material.

I take it this is all correct?

If it is, what I'm wondering is how I would be able to find the dimensions that would use the maximum amount of material? I only got one critical point when differentiating and that turned out to be a local minmum so how is it possible to find the maximum amount of material used?
• Apr 13th 2011, 09:20 AM
mtpastille
The math doesn't lie: if there is no maximum, then it is theoretically possible to use an infinite amount of material to build a box of that volume, unless there are any bounding values for the individual dimensions.
• Apr 13th 2011, 10:49 AM
HallsofIvy
For example, if you were to make the length of a side of the base, w, equal to 0.0001, then the height would be $\frac{32000}{0.000001}= 32000000000$ and so the total surface area would be 0.000001+ 4(32000000000)(0.001)= 128000000.000001. By making the base smaller and smaller, you can make the height, and so the surface area, arbitrarily large.

More generally, the derivative for f(w) you got was $f'= 2w- \frac{128000}{w^2}= \frac{2w^3- 128000}{w^2}= 2\frac{w^3- 64000}{w^2}= 2\frac{(w- 40)(w^2+ 40w+ 1600)}{w^2}$. Yes, f'(40)= 80- 80= 0. If w is larger than 50 then w- 40>0 and the other numbers are clearly positive so f'> 0. That is, f is increasing for all w> 40 and so there is no maximum surface area.
• Apr 13th 2011, 12:51 PM
nukenuts
That's pretty crazy hehe, thanks for clearing it up for me.