From a fixed point $A$ on the circumference of a circle of radius $a$,let the perpendicular $AY$ fall on the tangent at a point $P$ on the circle,prove that the greatest area which the $\Delta APY$ can have is $3\sqrt3\frac{a^2}{8}$sq units.
I cannot solve this question.How to determine the area of $\Delta APY$ which is to be maximized by differentiating.Please help me.
Hint: place the center of the circle at the origin and the point $A$ at $(-a,0)$. An arbitrary point $P$ on the circle can be described as $(a\cos t, a\sin t)$, $t\in[0,2\pi]$. Express the area of the corresponding right angle triangle $APY$ as a function of $t$ and maximize it.