Let $(h, k)$ be the point on ellipse through which tangent is passing. Equation of tangent at
$$
(h, k)=\frac{x h}{16}+\frac{y k}{81}=1
$$
at $y=0, x=\frac{16}{h}$
at $x=0, y=\frac{81}{k}$
Area of $\mathrm{AOB}=\frac{1}{2} \times\left(\frac{16}{h}\right) \times\left(\frac{81}{k}\right)=\frac{648}{h k}$
$$
\mathrm{A}^2=\frac{(648)^2}{h^2 k^2}
$$
$(h, k)$ must satisfy equation of ellipse
$$
\begin{aligned}
&\frac{h^2}{16}+\frac{k^2}{81}=1 \\
&h^2=\frac{16}{81}\left(81-k^2\right)
\end{aligned}
$$


Putting value of $h^2$ in equation (1)
$$
\mathrm{A}^2=\frac{81(648)^2}{16 \times k^2\left(81-k^2\right)}=\frac{\alpha}{81 k^2-k^4}
$$
differentiating w.r. to $k$

$$
\begin{aligned}
&2 \mathrm{AA}^{\prime}=\alpha\left(\frac{-1}{81 k^2-k^4}\right)\left(162 k-4 k^3\right) \\
&2 \mathrm{AA}^{\prime}=-2 \mathrm{~A}\left(81 k-4 k^3\right) \\
&\Rightarrow \mathrm{A}^{\prime}=-81 k-4 k^3 \\
&\text { Put } \mathrm{A}^{\prime}=0 \\
&\Rightarrow 162 k-4 k^3=0, k\left(162-4 k^2\right)=0 \\
&\Rightarrow k=0, k=\pm \frac{9}{\sqrt{2}} \\
&\mathrm{~A}^{\prime \prime}=-\left(81-12 k^2\right)
\end{aligned}
$$
For both value of k, A" $=405>0$
Area will be minimum for $k=\pm \frac{9}{\sqrt{2}}$
$$
\begin{aligned}
&h^2=\frac{16}{81}\left(81-k^2\right)=8 \\
&h=\pm 2 \sqrt{2}
\end{aligned}
$$
Area of triangle $\mathrm{AOB}=\frac{648 \times \sqrt{2}}{2 \sqrt{2} \times 9}=36$
sq unit