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

 Aug 29, 2019
 #1
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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

             

 

 

cool cool cool

 Aug 29, 2019

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