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FInd the maximum value of sqrt(x) - x, for x > 0.

 Jun 16, 2020
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Let \(t = \sqrt x\)

\(t^2 = x\)

 

So the objective function is just \(t - t^2\).

 

Completing the square yields \(t - t^2 = \dfrac14 - \left(t - \dfrac12\right)^2\)

 

So, when \(t = \dfrac12\), the maximum value of the objective function is attained, and the value is \(\boxed{\dfrac14}\)

 Jun 17, 2020

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