Find the maximum value of \(2x + 2\sqrt{x(1-x)}\) when \(0 \leq x \leq 1.\)
+128473 f(x) = 2x + 2[ x ( 1 - x) ]^(1/2)
f'(x) 2x + 2 [ x - x^2]^(1/2) take the derivative
f '(x) = 2 + [ x - x^2]^(-1/2) (1 - 2x) set this to 0
2 = (2x - 1) * [x - x^2]^(-1/2)
2 (x - x^2)^(1/2) = 2x - 1 square both sides
4 (x - x^2) = 4x^2 - 4x + 1
4x - 4x^2 = 4x^2 - 4x + 1 simplify as
8x^2 - 8x + 1 = 0
8 (x^2 - x + 1/8) = 0
x^2 - x + 1/8 = 0
x^2 - x + 1/4 = -1/8 + 1/4
(x - 1/2)^2 = 1/8 take the square root of both sides
x - 1/2 = ±√(1/8)
x = ±√(1/8) + 1/2
The positive root gives the x value of the maximum as √[1/8] + 1/2 ≈ .3004
And the max value is ≈ 2.41421
