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zmudz
 one year ago
Find the value of \(x\) that maximizes
\(f(x) = \log (20x + 12\sqrt{x}).\)
Note: there may not be a maximum value at all.
zmudz
 one year ago
Find the value of \(x\) that maximizes \(f(x) = \log (20x + 12\sqrt{x}).\) Note: there may not be a maximum value at all.

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BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2have you tried getting the derivative of f(x)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's an approach that doesn't rely on calculus. Recall that \(\log z\) is defined for \(z>0\) (where \(z\in\mathbb{R}\)). In this case, \(z=12\sqrt x20x\). You have \(z>0\) whenever \(12\sqrt x20x>0\), or \(4\sqrt x(35\sqrt x)>0\). The factor \(4\sqrt x\) will always be positive, so you need to require that \(35\sqrt x>0\), i.e. \(x<\dfrac{9}{25}\). And because you're dealing with square roots, you're forced to work with \(x\) such that \(0<x<\dfrac{9}{25}\). Anyway, one thing you might know about the logarithm is that it attains its extrema at the same points as its arguments. In other words, \(\log z\) will be max/min'd when \(z\) is max/min'd. \[\begin{align*} 12\sqrt x20x&=12t20t^2&\text{where }t=\sqrt x\\[1ex] &=\frac{1}{20}\left(t^2\frac{3}{5}t\right)\\[1ex] &=\frac{1}{20}\left(t^2\frac{3}{5}t+\frac{9}{100}\frac{9}{100}\right)\\[1ex] &=\frac{1}{20}\left(\left(t\frac{3}{10}\right)^2\frac{9}{100}\right)\\[1ex] &=\frac{1}{20}\left(t\frac{3}{10}\right)^2+\frac{9}{2000}\\[1ex] &=\frac{1}{20}\left(\sqrt x\frac{3}{10}\right)^2+\frac{9}{2000}\\[1ex] \end{align*}\] This is really a parabola in disguise. What I did here was complete the square to write it in vertex form. The vertex of any parabola is its extremum; since the coefficient of the leading term is negative, the parabola "opens downward", which means its extremum is a maximum.
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