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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 16:42:48
![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](/uploads/image/z/2605356-36-6.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E4%B8%BA%E6%AD%A3%E5%AE%9E%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BB%E5%BC%8Fx%5E2-2x%2B4y%5E2%3D0%2C%E6%B1%82xy%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.%E4%BB%A4xy%3Dp%E5%B7%B2%E7%9F%A5x%2Cy%E4%B8%BA%E6%AD%A3%E5%AE%9E%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BB%E5%BC%8Fx%5E2-2x%2B4y%5E2%3D0%2C%E6%B1%82xy%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.%E4%BB%A4xy%3Dpy%3Dp%2Fxx%5E4-2x%5E3%2B4p%5E2%3D04p%5E2%3D2x%5E3-x%5E4%3Dx%5E3%282-x%29%3D27%2A%28x%2F3%29%5E3%2A%282-x%29%26lt%3B%3D27%2A%282%2F4%29%5E4%3D27%2F16p%5E2%26lt%3B%3D27%2F6)
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.