CEMC Fermat, 2022

Suppose that the original can has radius $r$ cm and height $h$ cm.

Since the surface area of the original can is $300\text{ cm}^2$, then $2\pi r^2 + 2\pi r h = 300$.

When the radius of the original can is doubled, its new radius is $2r$ cm, and so an expression for its
surface area, in $\text{cm}^2$, is
$2\pi (2r)^2 + 2\pi (2r) h$ which
equals $8\pi r^2 + 4\pi rh$, and so
$8 \pi r^2 + 4\pi rh = 900$.

When the height of the original can is doubled, its new height is $2h$ cm, and so an expression for its
surface area, in $\text{cm}^2$, is
$2\pi r^2 + 2\pi r (2h)$ which equals
$2\pi r^2 + 4\pi rh$.

Multiplying $2\pi r^2 + 2\pi r h = 300$ by 3, we obtain $6\pi r^2 + 6\pi r h = 900$.

Since $8 \pi r^2 + 4\pi rh = 900$, we
obtain $$\begin{align*} 6\pi r^2 + 6\pi r h & = 8 \pi r^2 + 4\pi rh \\ 2\pi r h & = 2\pi r^2 \\ \pi r h & = \pi r^2\end{align*}$$ Since $2\pi r^2 + 2\pi r h = 300$ and $\pi rh = \pi r^2$, then $2\pi r^2 + 2\pi r^2 = 300$ and so $4\pi r^2 = 300$ or $\pi r^2 = 75$.

Since $\pi rh = \pi r^2 = 75$, then
$2\pi r^2 + 4\pi rh = 6 \cdot 75 = 450$, and so the surface area of the cylinder with its height
doubled is $450\text{ cm}^2$.