# Maximization Problem

• May 10th 2008, 08:56 PM
Hibijibi
Maximization Problem
Just wondering the basic process into solving this one.

A square is to be cut from each corner of a piece of paper which is 8cm by 10cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume?
• May 10th 2008, 09:15 PM
Hibijibi
Nvm I solved it
• May 10th 2008, 09:17 PM
TheEmptySet
Quote:

Originally Posted by Hibijibi
Just wondering the basic process into solving this one.

A square is to be cut from each corner of a piece of paper which is 8cm by 10cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume?

Attachment 6274

so the volume of the box will be

$\displaystyle V=l\cdot w \cdot h =(10-2x)(8-2x)x=80x-36x^2+4x^3$

taking the derivative we get the max and min points

answer is below (if you scroll down.)

$\displaystyle \frac{dV}{dx}=80-72x+12x^2$

using the quadratic formula we get

$\displaystyle x=\frac{-(-72) \pm \sqrt{(-72)^2}-4(12)(80)}{2(12)}=3 \pm \frac{\sqrt{21}}{3}$
• May 10th 2008, 09:20 PM
Hibijibi
Forgot about quadratic, I just graphed and found 0s that way. Also you forgot to use (10-2x)(8-2x)(x), instead of (10-x)(8-x)(x). You get different answers if you forget the 2x.
• May 10th 2008, 09:22 PM
TheEmptySet
Quote:

Originally Posted by Hibijibi
Forgot about quadratic, I just graphed and found 0s that way. Also you forgot to use (10-2x)(8-2x)(x), instead of (10-x)(8-x)(x). You get different answers if you forget the 2x.

Yep. They are in my diagram but not my equations, I will fix it.

Thanks