Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 10

Suppose that a box with an open top is to be constructed from a square piece of cardboard, 3 $ft$ wide, by cutting out a square from each of the four corners. Find the largest volume that such a box can have.



The volume = $x (3-2x)(3-2x) = x(3-2x)^2$
If we take the derivative of the volume by using Product and Chain Rule,

$
\begin{equation}
\begin{aligned}
v' &= x\left( 2(3-2x)(-2) \right) + (3-2x)^2 (1)\\
\\
v' &= (3-2x) [-4x + (3-2x)]\\
\\
v' &= (3-2x) [3-6x]\\
\\
v' &= 9 - 18x - 6x + 12x^2\\
\\
v' &= 12x^2 - 24x + 9
\end{aligned}
\end{equation}
$

when $v' = 0$,
$0 = 12x^2 - 24x + 9$
Using Quadratic Formula, we have
$\displaystyle x = \frac{3}{2}ft \text{ and } x = \frac{1}{2} ft$

Let us determine on which dimension will the volume be larger.

$
\begin{equation}
\begin{aligned}
\text{so when } x &= \frac{3}{2} ft, &&& \text{when } x &= \frac{1}{2} ft,\\
\\
v \left( \frac{3}{2} \right) &= \frac{3}{2} \left(3-2 \left(\frac{3}{2}\right) \right)^2 &&& v \left( \frac{1}{2} \right) &= \frac{1}{2} \left(3-2 \left(\frac{1}{2}\right) \right)^2\\
\\
v \left( \frac{3}{2} \right) &= 0 ft^3 &&& v \left( \frac{1}{2} \right) &= 2ft^3
\end{aligned}
\end{equation}
$

Therefore, the olume will be largest at $\displaystyle x = \frac{1}{2}ft$