`A B` is a diameter of a circle and `C` is any point on the circle. Show that the area of ` A B C` is maximum, when it is isosceles.

Explanation

We have, AB=2r
and `angleACB=90^(@)` [since, angle in the semi-circle is always `90^(@)`]
Let AC=x and BC=y
`therefore (2r)^(2)=x^(2)+y^(2)`
`rArr y^(2)=4r^(2)-x^(2)`
`rArr y=sqrt(4r^(2)-x^(2))` ……………………(i)
Now, area of `DeltaABC, A=1/2 xx x xx y`
`=1/2 xx x xx (4r^(2)-x^(2))^(1//2)` [using Eq. (i)]
Now, differentiating both sides w.r.t. x, we get
`(dA)/(dB) = 1/2[x.1/2(4r^(2)-x^(2))^(-1/2). (0-2x)+(4r^(2)-x^(2))^(1//2).1]`
`=1/2[(-2x)^(2)/(2sqrt(4r^(2)-x^(2)))+(4r^(2)-x^(2))^(1//2)]`
`1/2[(-x^(2))/(sqrt(4r^(2)-x^(2)))+sqrt(4r^(2)-x^(2))]`
`=1/2[(-x^(2)+4r^(2)-x^(2))/(sqrt(4r^(2)-x^(2)))]=1/2[(-2x^(2)+4r^(2))/(sqrt(4r^(2)-x^(2)))]`
`rArr (dA)/(dx) = [(-x^(2)+2r^(2))/(sqrt(4r^(2)-x^(2)))]`
Now, `(dA)/(dx) =0`
`rArr -x^(2)+2r^(2)=0`
`rArr r^(2)+1/2x^(2)`
`rArr r=1/sqrt(2)x^(2)`
`therefore x=rsqrt(2)`x

Again, differentiating both sides, w.r.t. x, we get
`(d^(2)A)/(dx^(2)) =(sqrt(4r^(2)-x^(2)).(-2x)+(2r^(2)-x^(2)).1/2(4r^(2)-x^(2))^(1//2)(-2x))/(sqrt(4r^(2)-x^(2))^(2))`
`=(-4x(sqrt(4r^(2)-x^(2)))+(2r^(2)-x^(2))(-2x))/(2.(4r^(2)-x^(2))^(3//2))`
`=(-4x(4r^(2)-x^(2))+(2r^(2)-x^(2)).(-2x))/(2.(4r^(2)-x^(2))^(3//2))`
`=(-16xr^(2)+4x^(3)+(2r^(2)-x^(2))(-2x))/(2.(4r^(2)-x^(2))^(3//2))`
`(d^(2)A)/(dx^(2))_(x=rsqrt(2))=(-16.rsqrt(2).r^(2)+4.(rsqrt(2))^(3)+[2r^(2)-(rsqrt(2))^(2)].(-2.rsqrt(2)))/(2.(4r^(2)-2r^(2))^(3//2))` `[therefore x=rsqrt(2)]`
`=(16.sqrt(2).r^(3)+8sqrt(2)r^(3))/(2(2r^(2))^(3//2))=(8sqrt(2)r^(2)[r-2])/(4r^(3))`
`=(-8sqrt(2)r^(3))/(4r^(3))=-2sqrt(2) lt0`
For `x=rsqrt(2)`, the area of triangle is maximum.
For `x=rsqrt(2)`, `y=sqrt(4r^(2)-(rsqrt(2))^(2))=sqrt(2r^(2))=rsqrt(2)`
Since, `x=rsqrt(2)=y`
Hence, the triangle is isosceles.