Find the [area](https://artofproblemsolving.com/wiki/index.php/Area) of the region enclosed by the [graph](https://artofproblemsolving.com/wiki/index.php/Graph) of $|x-60|+|y|=\left|\frac{x}{4}\right|.$
[1987 AIME-4.png](https://artofproblemsolving.com/wiki/index.php/File:1987_AIME-4.png)
Since $|y|$ is [nonnegative](https://artofproblemsolving.com/wiki/index.php/Nonnegative), $\left|\frac{x}{4}\right| \ge |x - 60|$. Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$. Thus, $48 \le x \le 80$. The [maximum](https://artofproblemsolving.com/wiki/index.php/Maximum) and [minimum](https://artofproblemsolving.com/wiki/index.php/Minimum) y value is when $|x - 60| = 0$, which is when $x = 60$ and $y = \pm 15$. Since the graph is [symmetric](https://artofproblemsolving.com/wiki/index.php/Symmetry) about the y-axis, we just need [casework](https://artofproblemsolving.com/wiki/index.php/Casework) upon $x$. $\frac{x}{4} > 0$, so we break up the condition $|x-60|$:

$x - 60 > 0$. Then $y = -\frac{3}{4}x+60$.
$x - 60 < 0$. Then $y = \frac{5}{4}x-60$.
The area of the region enclosed by the graph is that of the quadrilateral defined by the points $(48,0),\ (60,15),\ (80,0), \ (60,-15)$. Breaking it up into triangles and solving or using the [Shoelace Theorem](https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem), we get $2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}$.