Welcome to NRICH.

 
Differentiation for open box problem
By Norowi Hamid on February 19, 1998:

Dear mathematician,

Please show me how to solve this problem through differentiation method.

An open tank is constructed, with a square base and vertical sides to hold 32 cubic metres of water. Find the dimensions of the tank if the area of sheet metal used to make it if it is to have a minimum value.

Thank You
Norowi Hamid

By Sujata:

Dear Norowi,

It can be shown that the minimum surface area for a given volume occurs when the shape is spherical. So, you are looking for a tank that is as close to being spherical as possible: hence it is cubic. i.e. all the sides are x. The volume is x3 which is 32, or
x3 = 32
therefore, x = cuberoot (32)


If you really want to use differentiation, I would do the following:

Let the side of the base be x, and the height of the tank y.

Volume of tank = yx² = 32 .............. (1)
Surface Area = 2x² + 4xy = A .............. (2)

using (1) and (2) we have:
A = 2x² + 4(32/x)

To find the dimensions x and y such that area A is minimised, differentiate A w.r.t. x:
dA/dx = 4x - 4.(32/x² ) = 0
or,
4 x³ = 4.(32)
or,
x³ = 32 , or x = cube root of 32.

Using (1) you can find y, once you know x. It will turn out that y=x.

cheers,

Sujata.

By David Sanders on February 24, 1998:

I agree with Sujata's answer provided you take the tank as being closed, i.e. with a bottom and a top made out of metal.

If not, i.e. you just have a bottom, then:
volume V = yx2 = 32
surface area A = x2 + 4xy (not 2x2 + 4xy)

So A = x2 + 4(32/x)

so dA/dx = 2x - 128/x2

which is 0 when x3 = 64, i.e. x=4.

Since this looks to me exactly like an A-level homework question, I would expect a nice answer!

David.