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#?

1 Answer
Nov 10, 2017

#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]}