## 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.

have you tried getting the derivative of f(x)?

2. anonymous

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.