Length of rectangle $=10 \cos \theta$ and
breadth of rectangle $=8 \sin \theta$
$\therefore$ Area of rectangle $=(10 \cos \theta)(8 \sin \theta)=40(\sin \theta)$
Maximum area will occur when $\sin 2 \theta=1$
$$
\begin{aligned}
& \therefore \sin 2 \theta=\sin \frac{\pi}{2} \quad \Rightarrow \theta=\frac{\pi}{4} \\
& \therefore P=\left(\frac{5}{\sqrt{2}}, \frac{4}{\sqrt{2}}\right) \Rightarrow \text { Dimensions of rectangle are } 5 \sqrt{2}, 4 \sqrt{2}
\end{aligned}
$$