From each corner of a square piece of sheet metal $18$ ${\rm{cm}}$ on a side, we remove a small square and turn up the edges to form an open box. What are the dimensions of the box with the largest volume?

Solution

Step 1 The quantity to be maximized is the volume of the box; we denote it by $V$. We denote the length of each side of the small squares by $x$ and the length of each side after the small squares are removed by $y,$ as shown in Figure 54. Both $x$ and $y$ are in centimeters.

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Figure 54 $y=18 - 2x$

Step 2 Then $y=( 18-2x) \ {\rm{cm}}$. The height of the box is $x \ {\rm{cm}}$, and the area of the base of the box is $y^{2}\ {\rm{cm}}^{2}$. So, the volume of the box is $V=x{y^{2}}\ {\rm{cm}}^{3}.$

Step 3 To express $V$ as a function of one variable, we substitute $y=18-2x$ into the formula for $V$. Then the function to be maximized is $$V=V(x) =x({18-2x})^{2}$$

Since both $x\geq 0$ and $18-2x\geq 0,$ we find $x\leq 9$, meaning the domain of $V$ is the closed interval $[ 0,9] .$ (All other numbers make no physical sense—do you see why?)

Step 4 To find the value of $x$ that maximizes $V$, we differentiate $V$ and find the critical numbers: $${{V^\prime }(x)}=2x({18-2x})({-2})+({18-2x})^{2}=({18-2x})({18-6x})$$

Now, we solve $V^\prime (x)=0$ for $x$. The solutions are $$x=9\qquad \hbox{or}\qquad x=3$$

The only critical number in the open interval $(0,9)$ is $3$. We evaluate $V$ at $3$ and at the endpoints $0$ and $9$. $$V(0)=0\qquad V(3)=3(18-6)^{2}=432 \qquad V(9)=0$$

The maximum volume is $432\ {\rm{cm}}^{3}$. The box with the maximum volume has a height of $3\ {\rm{cm}}$. Since $y=18-2(3) =12\ {\rm{cm}}$, the base of the box measures $12\ {\rm{cm}}$ by $12\ {\rm{cm}}$.