求曲线X^3+Y^3-XY=1(X>=0,Y>=0)上点到原点的最长和最短距离
人气：121 ℃ 时间：2019-10-10 03:20:13
解答
限制条件：x^3+y^3-xy-1=0,x>=0,y>=0目标函数：x^2+y^2运用拉格朗日乘数方法设f(x,y)=x^2+y^2 + k*(x^3+y^3-xy-1)df/dx=0df/dy=0df/dk=0（d为偏导）得2x+3k*x^2-k*y=02y+3k*y^2-k*x=0x^3+y^3-xy-1=0解起来好...
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