Algebra Frenzy - Casework 1

$\frac{\partial f(a,x)}{\partial a}=-\frac{2 \log (x)}{\log (4)}$ informs us that the maximum occurs at $a=-2$ as the derivative is always negative. $\frac{\partial f(-2,x)}{\partial x}=\frac{\log (4 x)}{x \log ^2(4)}+\frac{\frac{\log (x)}{\log (4)}-3}{x \log (4)}+\frac{6}{x \log (4)}$ informs us the maximum occurs at $x=64$ as the derivative is always positive. $f(-2,64)=18$ .

No need for casework, or calculus - the following facts are all we need:

• $f$ is linear in the variable $a$ ; the coefficient of $a$ is negative
• $f$ is quadratic in $\log x$ ; the coefficient of $(\log x)^2$ is positive
• $\log x$ is monotonic and defined everywhere in the $x$ -interval

From these, it must be the case that the maximum value occurs at $a=-2$ and at one of the endpoints of the $x$ -interval. Checking, we find $f(-2,16)=9$ and $f(-2,64)=\boxed{18}$

Just to explain a bit more, if we think about $f$ as a surface plot, the three facts above tell us respectively:

• the cross-sections corresponding to fixed $x$ are straight lines that decrease as $a$ increases; so the maximum of $f$ occurs at the minimum value of $a$ , ie $-2$
• if we use a log-axis for $x$ , the cross-sections corresponding to fixed $a$ are "U-shaped" parabolas. There may or may not be a turning point in the interval, but if there is, it's a minimum; the maximum of $f$ must occur at either $x=16$ or $x=64$
• the transformation between $x$ and $\log x$ changes the shape of the cross-sections, but not drastically; also there are no "gaps" where $f$ is not defined to worry about