A rectangle is inscribed in a semi-circle of radius `r` with one of its sides on diameter of semi-circle. Find the dimensions of the rectangle so that its area is maximum. Find also the area.

Answer

length `= r sqrt2`, breadth `= (r sqrt2)/(2)`, area `= r^(2)` sq units

Explanation

Let ABCD be the rectangle of length 2x and breadth y, inscribed in a semicircle of radius r and centre O. Let `angleBOC = theta`
Then, `(y)/(r) = sin theta and (x)/(r) = cos theta`
`:.` area of the rectangle is given by
`A = 2xy = r^(2) sin 2 theta`
`rArr (dA)/(d theta) = 2r^(2) cos 2 theta and (d^(2)A)/(d theta^(2)) = - 4 r^(2) sin 2 theta`
Now, `(dA)/(d theta) = 0 rArr cos 2 theta = 0 rArr 2 theta = (pi)/(2) rArr theta = (pi)/(4)`
For this value of `theta`, we have `(d^(2)A)/(d theta^(2)) = - 4r^(2) lt 0`
`:.` area is maximum when `2x = 2r "cos"(pi)/(4) = r sqrt2 and y = r "sin " (pi)/(4) = (r sqrt2)/(2)`
Maximum area `= 2xy = (r sqrt2 xx r sqrt2)/(2) = r^(2)` sq units.