614. There is a square sheet of tin, the side of which $a=$ $=60 \mathrm{~cm}$. By cutting out equal squares from all its corners and folding up the remaining part, a box (without a lid) needs to be made. What should be the dimensions of the squares to be cut out so that the box has the maximum volume?
Solution. According to the condition, the side of the square $a=60$. Let the side of the squares cut out from the corners be denoted by $x$. The bottom of the box is a square with side $a-2 x$, and the height of the box is equal to the side $x$ of the cut-out square. Therefore, the volume of the box can be expressed by the function

$$
V=(a-2 x)^{2} x
$$

To find the value of $x$ at which the function will take the maximum value, we first transform the function and then investigate it for an extremum:

$$
\begin{gathered}
V=\left(a^{2}-4 a x+4 x^{2}\right) x ; \quad V=4 x^{3}-4 a x^{2}+a^{2} x \\
V^{\prime}=12 x^{2}-8 a x+a^{2} ; \quad 12 x^{2}-8 a x+a^{2}=0 \\
x=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{12}=\frac{4 a \pm \sqrt{4 a^{2}}}{12}=\frac{4 a \pm 2 a}{12} ; \quad x_{1}=\frac{a}{2}, \quad x_{2}=\frac{a}{6}
\end{gathered}
$$

Obviously, the value $x=a / 2$ does not meet the condition, as in this case the square would be cut into four equal parts and no box would be formed. Therefore, we investigate the function for an extremum at the critical point $x_{2}=a / 6$ :

$$
V^{\prime \prime}=24 x-8 a ; V^{\prime \prime}\left(\frac{a}{6}\right)=24 \frac{a}{6}-8 a=4 a-8 a=-4 a<0
$$

i.e., the maximum is achieved at $x=a / 6$. Thus, the side of the cut-out square should be equal to $x=a / 6$. In this specific case, when $a=60$ cm, we get $x=10$ cm.