2.
\(x\)、\(y\)為實數，且\(x^2+xy+y^2=6\)，試求\(x^2y+xy^2-x^2-2xy-y^2+x+y\)的最大值及最小值分別為多少？
[解答]
令\(\cases{x+y=u\cr xy=v}\)，其中\((x-y)^2\ge 0\)，\((x+y)^2\ge 4xy\)，\(u^2\ge 4v\)
\(x^2+xy+y^2=6\)，\((x+y)^2-xy=6\)，\(u^2-v=6\)，\(u^2=v+6\)，\(u^2\ge 4(u^2-6)\)，\(-2\sqrt{2}\le u \le 2\sqrt{2}\)
\(x^2y+xy^2-x^2-2xy-y^2+x+y=xy(x+y)-(x+y)^2+(x+y)=u(u^2-6)-u^2+u=u^3-u^2-5u\)
令\(f(u)=u^3-u^2-5u\)，\(f'(u)=3u^2-2u-5=0\)，\((u+1)(3u-5)=0\)，\(\displaystyle u=-1,\frac{5}{3}\)
\(f(-2\sqrt{2})=-8-6\sqrt{2}\)，\(f(-1)=3\)，\(\displaystyle f(\frac{5}{3})=-\frac{175}{27}\)，\(f(2\sqrt{2})=-8+6\sqrt{2}\)
最大值\(3\)，最小值\(-8-6\sqrt{2}\)
