Find the maximum value of \(2x + 2\sqrt{x(1-x)}\) when \(0 \leq x \leq 1.\)
+128474 2x + 2√ [ x ( 1-x) ] =
2x + 2 [ x - x^2}^(1/2) take the derivative and set to 0
2 + 2(1/2) [ x-x^2]^(-1/2) ( 1-2x) = 0
2 + (1-2x)
____________ = 0
√ (x-x^2)
2 √ [x- x^2 ] + 1 - 2x = 0
2 √ [ x - x^2] = 2x -1 square both sides
4 [x - x^2] = 4x^2 - 4x + 1
4x - 4x^2 = 4x^2 - 4x + 1
8x^2 -8x + 1 = 0
Using the quadratic Formula
x = 8 ± √ [ 64 - 32] 8 ± √ 32 8 ± 4√2 2 ± √2
______________ = __________ = ________ = _______ ≈ .853 or .146
16 16 16 4
When x = 2 - √2 , 2x + 2√ [ x ( 1-x) ] = 1
_____
4
When x = 2 + √2
_______ , 2x + 2√ [ x ( 1-x) ] = 1 + √2 ....and this is the max
4
