实数a.b.c满足x+y+z=5,xy+yz+zx=3,求z的最大值
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实数a.b.c满足x+y+z=5,xy+yz+zx=3,求z的最大值
实数a.b.c满足x+y+z=5,xy+yz+zx=3,求z的最大值
实数a.b.c满足x+y+z=5,xy+yz+zx=3,求z的最大值
x+y=5-z,(x+y)²=(5-z)²,(x+y)²/4>=xy,(5-z)² /4>=xy
xy+yz+zx=3, xy=3-z(x+y)=3-z(5-z)
(5-z)² /4>=3-z(5-z)
3z²-10z-13<=0
-1<=z<=13/3
z的最大值为13/3
z=5-(x+y)
3=xy+(x+y)z=xy+(x+y)[5-(x+y)]=xy+(x+y)*5-(x+y)^2<=(x+y)^2/4+(x+y)*5-(x+y)^2
=(x+y)*5-3(x+y)^2/4
3(...
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z=5-(x+y)
3=xy+(x+y)z=xy+(x+y)[5-(x+y)]=xy+(x+y)*5-(x+y)^2<=(x+y)^2/4+(x+y)*5-(x+y)^2
=(x+y)*5-3(x+y)^2/4
3(x+y)^2-20*(x+y)+12<=0
[3(x+y)-2][(x+y)-6]<=0
2/3<=x+y<=6
当带入z=5-(x+y)
有-1<=z<=5-2/3=13/3
所以z的最大值为13/3
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因为x+y+z=5,xy+yz+zx=3故有
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=25
x^2+y^2+z^2=19
且x^2>=0,y^2>=0
z^2=19-x^2+y^2<=19
故 MAX z=√19
f(x,y,z)=z+a(x+y+z-5)+b(xy+yz+zx-3)
f`x=a+b(y+z)=0
f`y=a+b(x+z)=0
y+z=x+z
x=y
f`z=1+a+b(x+y)=0
2x+z=5
x²+2xz=3
x²+2x(5-2x)=3
(x-3)(-3x+1)=0
x=3 x=1/3
z=-1 z=13/3
z的最大值13/3