Consider a right circular cylinder with radius $r$ and height $m$. There exists another right circular cylinder that is not congruent to the given one, but has the same volume and surface area as the given one. Determine the dimensions of the latter cylinder in general, and specifically, if $r=8 \text{~cm}, m=48 \text{~cm}$.

In what special case will the second cylinder be identical to the first one?
The surface area of the given right circular cylinder is $F=2 r^{2} \pi+2 r \pi m$ and its volume is $V=r^{2} \pi m$.

Let the radius of the other right circular cylinder be $r_{1}$, and its height be $m_{1}$, so its surface area and volume are $F_{1}=2 r_{1}^{2} \pi+2 r_{1} \pi m_{1}$ and $V_{1}=r_{1}^{2} \pi m_{1}$.

According to the problem,

$$
2 r^{2} \pi+2 r \pi m=2 r_{1}^{2} \pi+2 r_{1} \pi m \quad \text { and } \quad r^{2} \pi m=r_{1}^{2} \pi m_{1}
$$

or

$$
\begin{gathered}
r^{2}+m r=r_{1}^{2}+m_{1} r_{1} \\
r^{2} m=r_{1}^{2} m_{1}
\end{gathered}
$$

The system of equations (1) and (2) needs to be solved for the unknowns $r_{1}$ and $m_{1}$. From (1),

$$
m_{1}=\frac{r^{2}+m r-r_{1}^{2}}{r_{1}}
$$

From (2),

$$
m_{1}=\frac{m r^{2}}{r_{1}^{2}}
$$

Therefore:

$$
\frac{m r^{2}}{r_{1}^{2}}=\frac{r^{2}+m r-r_{1}}{r_{1}}
$$

Multiplying by $r_{1}^{2}$ and rearranging:

$$
r_{1}^{3}-r^{2} r_{1}-m r r_{1}+m r^{2}=0
$$

It is obvious that the pair of roots $r_{1}=r$ and $m_{1}=m$ satisfies the system of equations (1) and (2).

Dividing equation (3) by the root factor $\left(r_{1}-r\right)$, we get the quadratic equation

$$
r_{1}^{2}+r r_{1}-m r=0
$$

from which

$$
r_{1}=\frac{-r \pm \sqrt{r^{2}+4 m r}}{2}
$$

From the nature of the problem, it follows that $r_{1}>0$, so the appropriate root is:

$$
r_{1}=\frac{\sqrt{r^{2}+4 m r}-r}{2}
$$

and thus, from (2),

$$
m_{1}=\frac{1}{r_{1}^{2}} \cdot m r^{2}=\frac{r+2 m+\sqrt{r^{2}+4 m r}}{2 m^{2} r} \cdot m r^{2}=\frac{r^{2}+2 m r+r \sqrt{r^{2}+4 m r}}{2 m}
$$

With the given numerical values:

$$
\begin{aligned}
& r_{1}=\frac{\sqrt{8^{2}+4 \cdot 48 \cdot 8}-8}{2 m^{2} r}=\frac{\sqrt{1600}-8}{2}=16 \mathrm{~cm} \\
& m_{1}=\frac{m r^{2}}{r_{1}^{2}}=\frac{48 \cdot 8^{2}}{16^{2}}=12 \mathrm{~cm}
\end{aligned}
$$

If the second cylinder is identical to the first, then

$$
r=\frac{\sqrt{r^{2}+4 m r}-r}{2}, \text { or } 3 r=\sqrt{r^{2}+4 m r}
$$

Squaring and rearranging, we get $2 r^{2}-m r=0$, from which, apart from the trivial root $r=0$,

$$
r=\frac{m}{2}
$$

Therefore, the second cylinder is identical to the first if the radius of the given cylinder is half its height, i.e., the cylinder is equilateral.