Question

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius `a` is a square of side `sqrt(2)a` .

Answer 1

Let ABCD be the rectangle inscribed in a circle of radius a and with centre O. Join OC. Let `angleCOX = theta`. Then, the coordinates of C are `(a cos theta, a sin theta)`.
`:. OM = a cos theta and MC = a sin theta`
`BC = 2MC = 2 a and CD = 2OM = 2 a cos theta`
`:. P = 2a (cos theta + sin theta)`
So, `(dP)/(d theta) = 2a (-sin theta + cos theta) and (d^(2)P)/(d theta^(2)) = -2a (cos theta + sin theta)`
`(dP)/(d theta) 0 rArr theta = (pi)/(4)` and at this value of `theta, (d^(2)P)/(d theta) lt 0`
So, P is maximum at `theta = (pi//4)`
Then, `BC = sqrt2 a = CD`