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  • 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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  1. BAdhi
    • one year ago
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    have you tried getting the derivative of f(x)?

  2. anonymous
    • one year ago
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    Here'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, \(z=12\sqrt x-20x\). You have \(z>0\) whenever \(12\sqrt x-20x>0\), or \(4\sqrt x(3-5\sqrt x)>0\). The factor \(4\sqrt x\) will always be positive, so you need to require that \(3-5\sqrt x>0\), 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. \[\begin{align*} 12\sqrt x-20x&=12t-20t^2&\text{where }t=\sqrt x\\[1ex] &=-\frac{1}{20}\left(t^2-\frac{3}{5}t\right)\\[1ex] &=-\frac{1}{20}\left(t^2-\frac{3}{5}t+\frac{9}{100}-\frac{9}{100}\right)\\[1ex] &=-\frac{1}{20}\left(\left(t-\frac{3}{10}\right)^2-\frac{9}{100}\right)\\[1ex] &=-\frac{1}{20}\left(t-\frac{3}{10}\right)^2+\frac{9}{2000}\\[1ex] &=-\frac{1}{20}\left(\sqrt x-\frac{3}{10}\right)^2+\frac{9}{2000}\\[1ex] \end{align*}\] 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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