11. Given $x+y=1$, for what values of real numbers $x, y$ does $\left(x^{3}+1\right)\left(y^{3}+1\right)$ achieve its maximum value.
Solve:
$\left(x^{3}+1\right)\left(y^{3}+1\right)=x^{3} y^{3}+x^{3}+y^{3}+1=x^{3} y^{3}+(x+y)^{3}-3 x y(x+y)+1$ $=x^{3} y^{3}-3 x y+2$, since $x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, let $t=x y \in\left(-\infty, \frac{1}{4}\right]$, $f(t)=t^{3}-3 t+2$, then $f^{\prime}(t)=3 t^{2}-3=3(t+1)(t-1)$,
thus $f(t)$ is monotonically increasing on $(-\infty,-1)$, and monotonically decreasing on $\left(-1, \frac{1}{4}\right)$, and $f(-1)=4$, so the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$ is 4.
Equality holds when $\left\{\begin{array}{l}x y=-1, \\ x+y=1\end{array} \Rightarrow x, y\right.$ are the two roots of the equation $z^{2}-z-1=0$, thus $x=\frac{1+\sqrt{5}}{2}, y=\frac{1-\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}, y=\frac{1+\sqrt{5}}{2}$.