已知x,y满足x的绝对值加y的绝对值小于或等于1,求z=x^2j+y^2的最大值和最小值?

Question

如题所述

Answer 1

lxl+lyl<=1
只需考虑lxl+lyl=1 即可
令lxl=cos^2 x lyl=sin^2 x
z=cos^4 x +sin^4 x
=(cos^2 x+sin^2 x) ^2 -2sin^2 x cos^2 x
=1-2sin^2 x cos^2 x
=1-2sin^2 x (1-sin^2x)
=1-2sin^2 x +2sin^4 x
=2(sin^4 x-sin^2 x +1/4) +1/2
=2(sin^2 x-1/4)^2 +1/2
当sin^2 x=1/2 时 最小值为 1/2
当sin^2x =1 时 最大值为1追问

已知x,y满足x的绝对值加y的绝对值小于或等于1,求z=x^2-xy+y^2的最大值和最小值?

追答

0>=lxl+lyl>=1 只需考虑lxl+lyl=1 即可求得最大值

令lxl=cos^2 x lyl=sin^2 x
z=x^2-xy+y^2 ( -xy>=0 才能取到最大值)
不妨设x=cos^2x y=-sin^2 x
z=cos^4 x +sin^2 x cos^2 x +sin^4 x
=(cos^2 x+sin^2x)^2 -sin^2x cos^2x
=1-sin^2x cos^2 x
当lxl+lyl=0时 取最小值 z=0