1. ## mamimize rectangle area

An open box is made from a rectangular piece of cardboard by cutting equal squares from each corner and turning up the sides. Write the volume of the box as a function of x if the material is 20 inches by 12 inches. What is maximum volume of the box?

This was one problem that stumped me alot.
Thank you!!

2. x appears to be the length of the sides of the removed squares.
When the side lengths are cut out, the piece of cardboard looks like this and is then folded along the dotted lines to create an open topped box.

3. so what is the the volume of the box as a function of x and the maximum volume of the box?

thankss

4. The volume of the box is $length \times width \times height$
The maximum volume of the box will be at either a stationary point or at an endpoint.

5. After we cut the squares of length x, the base of the box will be a rectangle with sides 20-2x and 12-2x (because we've taken a square off at both ends of each side). The height of the box will simply be x. So the volume as a function of x would be V(x)=x(20-2x)(12-2x).

To find the max volume, ie. optimize V, differentiate and set the derivative to zero:

$V'(x)=(20-2x)(12-2x)-2x(12-2x)-2x(20-2x)=0$

$(10-x)(6-x)-x(6-x)-x(10-x)=0$ (divide by 4 everywhere)

$60-16x+x^2-6x+x^2-10x+x^2=0$

$3x^2-32x+60=0$