Question

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer 1

Let ABCD be a rectangle inscribed in the given circle of radius r having centre at O.

Let one side of the rectangle be c then the  other side

`= sqrt((2r)^2 - x^2)`

`= sqrt(4r^2 - x^2)`

(∴ ∠ADC = ∠ABC = 90°; an angle in the semi-circle).

Let A be the corresponding area of the rectangle.

`A = xsqrt(4r^2 - x^2), 0 < x < 2r`

⇒ `(dA)/dx = (x(-2x))/(2 sqrt(4r^2 - x^2)) + sqrt(4r^2 - x^2)`

`= (2(2r^2 - x^2))/sqrt(4r^2 - x^2)`

For maximum / minimum area

`(dA)/dx = (2r^2 - x^2)/(sqrt(4r^2 - x^2)) = 0`

⇒ `x = sqrt2r`          ...(∵ 0 < x < 2r) 

Now, 

`(d^2A)/dx^2 = (sqrt(4r^2 - x^2) (-4x) - (4r^2 - 2x^2) (1xx(-2x))/(2sqrt(4r^2 - x^2)))/((4r^2 - x^2))`

`= ((4r^2 - x^2) (-4x) + (4r^2 - 2x^2) x)/((4r^2 - x^2)^(3//2))`

`= (-12r^2x + 2x^3)/(4r^2 - x^2)^(3//2)`

`((d^2A)/dx^2)_(x = sqrt(2r))`

`= (-12r^2 (sqrt (2r)) + 2 (sqrt (2r))^3)/((2r^2)^(3//2)`

`= (4sqrt(2r^3) - 12 sqrt (2r^3))/(2 sqrt(2r^3))`

= 2 - 6 < 0

∴ Area is maximum at x = `sqrt(2r)`

∴ length of rectangle is `sqrt(2r)`

width of rectangle is `sqrt (4r^2 - x^2) = sqrt (2r)`

Hence the rectangle is a square of side `sqrt (2r)` for maximum area.