求解一道简单的基本不等式问题！（突然忘了）要详细步骤：若实数x>0 y>0 满足x+2y=1 则xy的最大值是？
\u82e5\u5b9e\u6570x>0,y>0,\u6ee1\u8db3x+2y=1,\u5219xy\u6700\u5927\u503c\u662f\uff1f\uff1f\u7528\u57fa\u672c\u4e0d\u7b49\u5f0f\u89e3\u3002\u3002\u3002\u6025a>=(x+\u6839\u53f7\uff08xy))/(x+2y)=(1+\u6839\u53f7(y/x))/(1+2y/x)=(1+t)/(1+2t^2)\uff0c\u7531\u9898\u610f\u5c31\u662f\u6c42\u51fd\u6570f(t)=(1+t)/(1+2t^2)\u5728t\u4f4d\u4e8e(0 +\u65e0\u7a77)\u4e0a\u7684\u6700\u5927\u503c\uff0c\u5176\u4e2dt=\u6839\u53f7(y/x)\u3002
\u4ee4(1+t)/(1+2t^2)=u\uff0c\u52192ut^2--t+u--1=0\uff0c\u8fd9\u4e2a\u5173\u4e8et\u7684\u4e8c\u6b21\u65b9\u7a0b\u8981\u60f3\u6709\u89e3\uff0c\u5fc5\u987b\u5224\u522b\u5f0f\u5927\u4e8e\u7b49\u4e8e0\uff0c
\u56e0\u6b641--8u(u--1)>=0\uff0c\u89e3\u5f97u<=\u30102+\u6839\u53f7(6)\u3011/4\uff0c\u56e0\u6b64a\u7684\u6700\u5c0f\u503c\u4e3a\u30102+\u6839\u53f7(6)\u3011/4\u3002
x=1-2y
xy=(1-2y)y
=-2y²+y
=-2(y²-y/2+1/16)+1/8
=-2(y-1/4)²+1/8
∴有最大值1/8
这个简单！xy=x（1-x）=-x^2+x,然后根据x的范围和二次函数的知识，可以求解吧！
Xy=y–4y²,下面就简单了
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