Answer: 480

Small Hint:

Solve for $|y|$ and determine where its right-hand side is nonnegative

Big Hint:

The boundary is a kite whose vertices occur at the breakpoints of the absolute values

Solution:

We need $|y|=\frac{|x|}{4}-|x-60|\geq0.$ This forces $48\leq x\leq80.$ At $x=48,$ $x=60,$ $x=80,$ the boundary values are respectively $y=0,$ $y=\pm15,$ $y=0.$ Hence the region is a kite with perpendicular diagonals $80-48=32$ and $30,$ so its area is $\frac12(32)(30)=480.$