Show that among all positive numbers `x` and `y` with `x^2+y^2=r^2` , the sum `x+y` is larger when `x=y=r//sqrt(2)` .

Explanation

`x^2+y^2=r^2`

`y=sqrt{r^2-x^2}`

Let `f=x+y=x+sqrt{r^2-x^2}`

`{df}/{dx}=1+1/{2sqrt{r^2-x^2}}xx(-2x)=0`

Put `{df}/{dx}=0`

`implies1+1/{2sqrt{r^2-x^2}}xx(-2x)=0`

`implies x=r/sqrt2`
`therefore {d^2f}/{dx^2}=-{ sqrt{r^2-x^2}+{x^2}/ sqrt{r^2-x^2}}/ {r^2-x^2}`

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