8. If $x$ and $y$ are real numbers with $(x+y)^{4}=x-y$, what is the maximum possible value of $y$ ?
Solution: $\quad 3 \sqrt[3]{2} / 16$
By drawing the graph of the curve (as shown), which is just a $135^{\circ}$ clockwise rotation and scaling of $y=x^{4}$, we see that the maximum is achieved at the unique point where $d y / d x=0$. Implicit differentiation gives $4(d y / d x+1)(x+y)^{3}=1-d y / d x$, so setting $d y / d x=0$ gives $4(x+y)^{3}=1$. So $x+y=1 / \sqrt[3]{4}=\sqrt[3]{2} / 2$, and $x-y=(x+y)^{4}=\sqrt[3]{2} / 8$. Subtracting and dividing by 2 gives $y=(\sqrt[3]{2} / 2-\sqrt[3]{2} / 8) / 2=3 \sqrt[3]{2} / 16$.