7. Let $x$ and $y$ be positive real numbers. What is the smallest possible value of $\frac{16}{x}+\frac{108}{y}+x y$ ?
(A) 16
(B) 18
(C) 24
(D) 30
(E) 36
7. Ans: E
Using Arithmetic Mean $\geq$ Geometric Mean, we have
$$
\begin{aligned}
\frac{1}{3}\left(\frac{16}{x}+\frac{108}{y}+x y\right) \geq \sqrt[3]{\frac{16}{x} \times \frac{108}{y} \times x y}=12 \\
\Longrightarrow \frac{16}{x}+\frac{108}{y}+x y \geq 36 .
\end{aligned}
$$

Equality is achieved when $\frac{16}{x}=\frac{108}{y}=x y=\frac{36}{3}=12$, which is satisfied when $x=\frac{4}{3}$ and $y=9$. Thus the smallest possible value is 36 .