2016-I-11

$\begin{array}{rcl}
\left( \dfrac{12}{16} \right)^3 & = & \dfrac{x}{ 444\pi + x} \\
\dfrac{27}{64} & = & \dfrac{x}{444\pi + x} \\
11988\pi + 27x & = & 64x \\
37x & = & 11988 \pi \\
x & = & 324 \pi
\end{array}$

Therefore, the final volume of milk in the vessel

$\begin{array}{cl}
= & 324\pi + 444 \pi \\
= & 768 \pi \text{ cm}^3
\end{array}$

$\begin{array}{rcl}
\dfrac{1}{3} \pi r^2 (16) & = & 768\pi \\
r^2 & = & 144 \\
r & = & 12
\end{array}$

Then, the final slant height of the milk in the vessel

$\begin{array}{cl}
= & \sqrt{12^2 + 16^2} \\
= & 20 \text{ cm}
\end{array}$

Hence, the final area of the wet curved surface of the vessel

$\begin{array}{cl}
= & \pi \times 12 \times 20 \\
= & 240 \pi \\
= & 753.982\ 236\ 9 \text{ cm}^2 \\
< & 800 \text{ cm}^2 \end{array}$

Therefore, I don’t agree with the claim.