Question

Show that the rectangle of the maximum perimeter which can be inscribed in the circle of radius 10 cm is a square of side `10sqrt2` cm.

Answer 1

`AB = 2 xx r cos theta`, `AD = 2r sin theta`

here r = 10

`AB = 20 cos theta, AD = 20 sin theta`

`A = 2rcos theta.2r sin theta`   {Here r = 10}

`A= 4r^2 cos theta.sin theta`

`= 2r^2. 2sin theta cos theta`

`= 2 xx 100 sin 2 theta`

`(dA)/(d theta) = 200 cos 2 theta . 2`

`(dA)/(d theta) = 0`

`=> cos 2 theta = 0`

`2 theta = pi/2`

`theta = pi/4`

`(dA^2)/(d theta^2)  = 400 (-2 sin 2 theta)`

`(dA^2)/(d theta^2) = -800 sin  pi/2 = - 800< 0`

So, Area is max, when `AB = 20 cos  pi/4 = 10sqrt2`

`AD = 20 cos pi/4 = 10sqrt2`

`:. AB = AD = 10sqrt2`