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` .

Explanation

Let `ABCD` be a rectangle inscribed in a circle of radius `a` with centre at `O` .

Let `AB =2x` and
`BC =2y` be the sides of the rectangle.

Then in right angled `△OAM`,
`AM^2+OM^2=OA^2`
`=>x^2+y^2=a^2`
`=>y=sqrt(a^2-x^2)`
Let `P` be the perimeter of rectangle, `ABCD` then

`P=4x+4y`
` ⟹P=4x+4sqrt(a^2-x^2)`
`=>(dp)/(dx)=4-(4x)/sqrt(a^2-x^2)`
For maximum or minimum value of P, we have
`(dp)/(dx)=0`
`=>4-(4x)/sqrt(a^2-x^2)=0`
`=>4=(4x)/sqrt(a^2-x^2)`
`=>sqrt(a^2-x^2)=x`
`=>a^2-x^2=x^2`
`=>2x^2=a^2`
`=>x=a/sqrt2`
Now,
`[(d^2p)/(dx^2)]_(x=a/sqrt2)=-(4a^2)/(a^2-a^2/2)^(3/2)`
`=(-8sqrt2)/a lt 0`
Thus, P is maximum when `x=a/sqrt2`.
putting `x=a/sqrt2` in (i), we get
`y=a/sqrt2`
Therefore, `x=y`
`⟹2x=2y`

Hence, P is maximum when rectangle is a square of side `2x= sqrt a`