困难的最大值

Canhuang

$x^2 + y^2 + x - 1 = 0$, 求 $(x - y + 1)^2 - xy(2x + y + 4)$ 的最大值.

hbghlyj

\begin{aligned}
& \max \left\{(x-y+1)^2-x y(2 x+y+4) \mid x^2+x+y^2-1=0\right\}=\frac{1}{54}(133+29 \sqrt{29}) \\
& \quad \text { at }(x, y)=\left(\frac{\sqrt{29}}{6}-\frac{1}{2},-\frac{2}{3}\right)
\end{aligned}