Question

Check whether the rectangle of maximum area which can be inscribed in a circle is a square?

Answer 1

Consider a circle of radius `r` centred on the origin, `O`

Let `ABCD` be a rectangle inscribed in the circle then:

`OA=OB=OC=OD=r`

Let `x=CD`, and. `y=BC` where `x,y gt 0`

Let us denote the area of the rectangle `ABCD` by `A`, then

`A = xy` ..... [A]

Also, the `triangle BCD` is right angled, so by Pythagoras:

`BD^2 = BC^2 + CD^2`
`:. (2r)^2 = y^2 + x^2`
`:. x^2 + y^2 = 4r^2`
`:. y = sqrt(4r^2 -x^2)` .... [B]

Substitution into [A] gives:

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

We want to maximise `A` wrt `x` so we examine values of `x` where:

`(dA)/dx = 0`

Differentiating wrt `x` we get;

`(dA)/dx = (x)(d/dx sqrt(4r^2 -x^2)) + (d/dx x)(sqrt(4r^2 -x^2))`
`" " = x(1/2 1/sqrt(4r^2 -x^2) * (-2x)) + sqrt(4r^2 -x^2)`
`" " = (4r^2 -x^2-x^2)/(sqrt(4r^2 -x^2))`
`" " = (4r^2 -2x^2)/(sqrt(4r^2 -x^2))`

At a maximum or minimum we have

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

Using [B] this gives:

`y = sqrt(4r^2 -2r^2)`
`\ \ = sqrt(2r^2)`
`\ \ = sqrt(2)r`

Thus we require `x=y`, ie a square

NB: We should validate that this value of `x` corresponds to a maximum using the second derivative test, but I will leave that fro the reader.