Open box of maximum volume

A square is cut from both ends of each side of the sheet, so every side gets $2x$ shorter. The base of the box therefore measures $(48 - 2x)$ by $(30 - 2x)$, and the folded-up sides give a height of $x$:

All three dimensions must be positive, otherwise there is no box. The height gives $x > 0$, and $30 - 2x > 0$ gives $x < 15$. The condition $48 - 2x > 0$ gives $x < 24$, but that already follows from $x < 15$. So:

The volume of a cuboid is length times width times height:

To make it easier to differentiate, expand it. First multiply out the two brackets:

Then multiply by $x$:

Differentiate term by term, using the power rule on each term:

Take out the common factor $12$ to get smaller numbers:

A maximum inside the domain can only occur where $V'(x) = 0$, so solve $x^2 - 26x + 120 = 0$. Here $a = 1$, $b = -26$ and $c = 120$, so work out the discriminant:

Its square root is $\sqrt{196} = 14$, so the quadratic formula gives:

The number $20$ is not in the domain $(0, 15)$, so inside the domain the derivative is zero only at $x = 6$. The roots also let you write the derivative in factorised form:

To tell whether $x = 6$ gives a maximum, look at the sign of $V'$ on each side of it. For every $x$ in the domain the factor $x - 20$ is negative, so the sign of the derivative depends only on the factor $x - 6$. Check this with two sample values:

So $V$ is increasing on the whole interval $(0, 6]$ and decreasing on the whole interval $[6, 15)$. This means that at $x = 6$ it has not just a local maximum but the greatest value on the whole domain.

Substitute $x = 6$ into the dimensions of the box:

Multiply them to get the volume: