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

Guest Apr 25, 2018
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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

 

 

 

cool cool cool

CPhill  Apr 25, 2018
edited by CPhill  Apr 25, 2018

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