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

Explanation

Let r be the radius of given circle. A rectangle ABCD is inscribed in the circle whose diagonal is AC= 2r.
Let ` AB = x and BC = y`
`:." In "Delta ABC`,
`AB^(2)+BC^(2)=AC^(2)`
` rArr x^(2) + y^(2) = (2r)^(2)`
` rArr y^(2) = 4r^(2) - x^(2)` ...(1)
Now, area of rectangle `A = x * y`
` rArr A = x sqrt(4r^(2)-x^(2))`
` rArr A^(2) = x^(2) (4r^(2)-x^(2))`
Let ` Z=A^(2)`
` rArr Z = 4r^(2)x^(2) - x^(4)`
` rArr (dz)/(dx) = 8r^(2)x-4x^(3)`
For maxima/minima
` (dZ)/(dx) = 0`
` rArr 8r^(2)x-4x^(3) = 0 rArr x^(2)=2r^(2)`
and ` ((d^(2)Z)/(dr^(2))) = 8r^(2) - 12x^(2)`
at ` x^(2) =2r^(2)`
` (d^(2)Z)/(dr^(2)) = 8r^(2) - 24r^(2) =- 16r^(2) lt 0 `
` rArr at x^(2) = 2r^(2) , Z` is maximum.
` rArr at x^(2) = 2r^(2), A^(2)` is maximum.
` rArr at x^(2) = 2r^(2), A` is maximum.
Now ` x^(2) = 2r^(2)`
` rArr 2x^(2) = 4r^(2) = x^(2)+y^(2)` [From equation (1)]
` rArr x^(2) = y^(2)`
` rArr x = y `
Therefore, the rectangle with maximum area, is a square.