# Optimization and Finding Dimensions

• Nov 16th 2010, 06:43 PM
Hoopsaholic31
Optimization and Finding Dimensions
Another problem I'm having a tough time with.

Find the dimensions of a box with a square bottom and no top that will hold exactly http://webwork.uoregon.edu/webwork2_...123f2a9831.png cubic inches and have the smallest possible surface area.

The answers are apparently 1.25992A by 1.25992A by 1.25992A/2, but I have no idea how to get there. Any help would be greatly appreciated. Thanks!
• Nov 16th 2010, 08:28 PM
Unknown008
The volume of the box is:

\$\displaystyle V = lbh\$

Since the bas is a square, let one side be x and the height be h, such that:

\$\displaystyle V = x^2h\$

This equals A^3;

\$\displaystyle A^3 = x^2h\$

Now, the Surface area is given by:

\$\displaystyle S.A = x^2 + 4xh\$

You have two equations now. Substitute h in the second equation and differentiate with respect to x and solve for 0, to get the value of x in those conditions (Don't forget that A^3 is a constant).

Then, substitute the value of x to get the value of h.
• Nov 17th 2010, 08:25 AM
Hoopsaholic31
Thanks, you are the bee's knees.