已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=p已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=py=p/xx^4-2x^3+4p^2=04p^2=2x^3-x^4=x^3(2-x)=27*(x/3)^3*(2-x)<=27*(2/4)^4=27/16p^2<=27/6
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已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=p已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=py=p/xx^4-2x^3+4p^2=04p^2=2x^3-x^4=x^3(2-x)=27*(x/3)^3*(2-x)<=27*(2/4)^4=27/16p^2<=27/6
已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=p
已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.
令xy=p
y=p/x
x^4-2x^3+4p^2=0
4p^2=2x^3-x^4=x^3(2-x)=27*(x/3)^3*(2-x)<=27*(2/4)^4=27/16
p^2<=27/64
p<=3√3/8
里面的27*(x/3)^3*(2-x)<=27*(2/4)^4怎么得来的?
已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=p已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值.令xy=py=p/xx^4-2x^3+4p^2=04p^2=2x^3-x^4=x^3(2-x)=27*(x/3)^3*(2-x)<=27*(2/4)^4=27/16p^2<=27/6
以上省略
4p²=2x³-x^4
=x³(2-x)
=(3·3·3)·(x/3)·(x/3)·(x/3)·(2-x)
≤27·[(x/3+x/3+x/3+2-x)/4]^4 (五元均值不等式)
=27·(2/4)^4
=27/16
→p²≤27/64
两边开方,即得p≤3√3/8.
所求最大值为p|max=3√3/8.