Question

Find the maximum value of `2x + 2sqrt{x(1-x)}` when `0 \leq x \leq 1`?

Original syntax:

Find the maximum value of `2x + 2sqrt{x(1-x)}` when `0 \leq x \leq 1`?

Answer 1

`sqrt(2)+1`

Explanation:

Let:

`f(x) = 2x+2sqrt(x(1-x))`

`color(white)(f(x)) = 2x+2sqrt(x-x^2)`

Then:

`f'(x) = 2+1/sqrt(x(1-x)) * (1-2x)`

`color(white)(f'(x)) = (2sqrt(x(1-x))+1-2x)/sqrt(x(1-x))`

The numerator of `f'(x)` is zero when:

`2sqrt(x(1-x)) = 2x-1`

Squaring both sides, this becomes:

`4(x(1-x)) = 4x^2-4x+1`

That is:

`4x-4x^2 = 4x^2-4x+1`

So:

`0 = 8x^2-8x+1 = 2(2x-1)^2-1`

So:

`(2x-1)^2 = 1/2`

So:

`2x-1 = +-sqrt(2)/2`

So:

`2x = 1+-sqrt(2)/2`

So:

`x = 1/2+-sqrt(2)/4`

We find:

`f(0) = 0`

`f(1/2-sqrt(2)/4) = 1`

`f(1/2+sqrt(2)/4) = sqrt(2)+1`

`f(1) = 2`

So the maximum value of `f(x)` on the interval `[0, 1]` is `sqrt(2)+1`

graph{2x+2sqrt(x(1-x)) [-0.2, 1.2, -0.4, 2.6]}