From a fixed point `A` on the circumference of a circle of radius `r ,` the perpendicular `A Y` falls on the tangent at `Pdot` Find the maximum area of triangle `A P Ydot`

Explanation

the figure AP=2r sin `theta`
`PY=2r sin theta cos theta = r sin 2 theta`
`AY = 2r sin theta sin theta`
`triangle` Area of `triangle APY=(1)/(2)PY.AY`
`=r^(2) sin^(2) theta sin 2 theta ,0lt theta lt pi//2`
`d triangle /d theta=r^(2)[sin^(2)2 theta+2cos 2 theta sin^(2) theta]`
`=r^(2)[4 sin^(2) theta cos^(2) theta+2 sin^(2) theta cos 2 theta]`
`=2r^(2) sin^(2) theta [4 cos^(2) theta-1]`
`d triangle /d theta= rArr sin theta =0, cos theta = pm 1//2`
Therefore the only ciritical point with in `(0,pi//2) is pi//3`. clearly
`(d triangle)/(d theta)_(pi//3-h) gt 0 and (d triangle)/(d theta)_(pi//3+h)lt0`
Thus `triangle` is maximum at `theta = pi//3`.Therefore the greatest area of such `triangle =3sqrt(3r^(2))//8`