MT【291】2元非齐次不等式

实数$x,y$满足$x^2+y^2=20,$求$xy+8x+y$的最大值___

法一:$xy\le\dfrac{1}{4}x^2+y^2,8x\le x^2+16,y\le\dfrac{1}{4}y^2+1,$故$xy+8x+y\le\dfrac{5}{4}(x^2+y^2)+17=42$
法二:$(xy+8x+y)^2\le (x^2+8^2+y^2)(y^2+x^2+1^2)=84*21=42^2$
法三:记$f(x,y,k)=xy+8x+y-k(x^2+y^2-20)$,令$f^{'}_x=0,f^{'}_y=0,f^{'}_k=0$得
$y+8-2kx=0,x+1-2ky=0,x^2+y^2-20=0$从而得到$x=4,y=2,k=\dfrac{5}{4} $,
故由拉格朗日配方法得
$4(xy+8x+y)-5(x^2+y^2-20)=-5(x-\dfrac{16+2y}{5})^2-\dfrac{21}{5}(y-2)^2+168\le 168$即$xy+8x+y\le 42$