883

已知 $x,y>0,x^3+y^3{\displaystyle -\frac{x}{4}-\frac{y}{4}=3}$, 求 $13x+y$ 的最大值.

由三元均值不等式得$$\begin{array}{rl}& x^3+{\left(\frac{3}{2}\right)}^3+{\left(\frac{3}{2}\right)}^3\ge 3\cdot \frac{9}{4}x,\\ & y^3+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^3\ge 3\cdot \frac{1}{4}y,\end{array}$$
相加得$$\begin{array}{r}{x}^3+y^3+7=3+\frac{1}{4}(x+y)+7\ge \frac{3}{4}(9x+y)\Rightarrow 13x+y\le 20,\end{array}$$
等号成立当且仅当 $(x,y)={\displaystyle (\frac{3}{2},\frac{1}{2})}$.