1.

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

Answer» Here, `x^2+y^2 = r^2`
`=> y = sqrt(r^2-x^2)`
Let `S = x+y`
`=> S = x+sqrt(r^2-x^2)`
For `S` to be largest, `(dS)/dx =0`.
`:. (dS)/dx = 1+1/sqrt(r^2-x^2)(-2x)`
`:. 1+1/(2sqrt(r^2-x^2))(-2x) = 0`
`=>x = sqrt(r^2-x^2)`
`=>x^2 = r^2-x^2`
`=>2x^2 = r^2`
`=>x = +-r/sqrt2`
As `x` is a positive number,
`:. x = r/sqrt2`.
Now, `y = sqrt(r^2-x^2) = sqrt(r^2-r^2/2) = r/sqrt2`
To further prove, that this is largest sum, we can take `(d^2S)/dx^2` at `x = r/sqrt2`.
It will come negative, which proves sum `S` is largest at these values.
So, `x+y` is largest when `x = y = r/sqrt2`.


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