minimizing the dimensions of a box help!

[Guest]

Here's the problem: A cardboard box without a lid is to have a volume of 32000 cm^3. Find the dimensions that minimize the amount of cardboard used.

We only have one example given to us to help us with this, the only problem with it is that the amount of material being used (essentially the surface area of it) is given to us instead of the volume. Here is the example:

A rectangular box without a lid is to be made from 12 m^2 of cardboard. Find the maximum colume of such a box.

Here it was easy to determine an equation for the volume after we determined the equation: (x as length, y as width, z as height) 2xz + 2yz + xy = 12. Any suggestion as to how I could come up with a similar equation with x, y, and z in it? Thank you to whomever assists with this.

 

[galactus]

I am going to use 32 instead of 32000. Just multiply your answers by 10.

The volume of the box is given by \(\displaystyle xyz=32\).........[1]

The surface area is given by \(\displaystyle xy+2xz+2yz\).......[2]

Solving [1] for z, we get \(\displaystyle z=\frac{32}{xy}\)

Sub into [2]: \(\displaystyle xy+\frac{64}{x}+\frac{64}{y}\).....[3]

Now, S is expressed as two variables.

Differentiate [3] with respect to x and y:

\(\displaystyle \frac{{\partial}S}{{\partial}x}=y-\frac{64}{x^{2}}\)

\(\displaystyle \frac{{\partial}S}{{\partial}y}=x-\frac{64}{y^{2}}\)

Therefore, the critical points are at:

\(\displaystyle y-\frac{64}{x^{2}}=0\).............. [4]

and

\(\displaystyle x-\frac{64}{y^{2}}=0\)............[5]

Solve [4] for y and sub into [5]

Solve for x. Back substitute to find y and z

To see that you have found a relative minimum, you can use the second partials test.

I doubt if you'll need it though. You will get two answers upon solving, one will be 0, the other the one you need.