• Oct 1st 2009, 02:01 PM
The D00D
What is the maximum volume possible from a piece of cardboard that is made into a cube, that measures 16 inches by 12 inches?

[___________ ]
[ __________ ]
[___________ ] 12 - 2x
[____________]
16in - 2x

• A Box with no top.

\$\displaystyle L = 12-2x\$
\$\displaystyle W = 16 - 2x\$
\$\displaystyle n = x \$

\$\displaystyle V = 172 - 32x - 24x + 4x^2(x)\$

\$\displaystyle V = 172x - 32x^2 - 24x^2 + 4x^2\$

\$\displaystyle 4x^3 - 24x^2 - 32x^2 + 172x\$

\$\displaystyle 4x^3 - 56x^2 +172x \$

\$\displaystyle 4x(x^2 - 14x + 43) \$
\$\displaystyle How to solve for X.\$

this is where I got stuck.

Thank you
• Oct 1st 2009, 02:06 PM
Matt Westwood
Quote:

Originally Posted by The D00D
Note: One with x#'s, are exponents, since I don't know to do an exponent on here

Thank you

Use ^ (it's <shift>6 on most keyboards) so x^2 becomes \$\displaystyle x^2\$.
• Oct 1st 2009, 03:09 PM
The D00D
Thank you for that.
• Oct 1st 2009, 09:36 PM
Matt Westwood
Okay so I've been looking at this question and I'm not sure I understand what it's asking, so I'm not sure how you got your initial equations.

But given your starting equations, I rather think you'll find that in your first step, 12 x 16 is a little bit more than 172.

Your end equation should be something like
\$\displaystyle 4x (x^2 - 14x + 48)\$