图解拉格朗日乘子法

所需积分/C币:10 2018-03-24 14:44:56 1.39MB PDF

在数学中的最优化问题中,拉格朗日乘数法方法可以将一个有n个变量与k个约束条件的最优化问题转换为一个解有n + k个变量的方程组的解的问题。本文以可视化的形式讲解拉格朗日乘子法,注意,虽然是英语,但是都很简单哦
y The work has already been done. Here I would like to rephrase the boundary problem using our new language. Find the absolute maximum and minimum of f(x,y)=x2+y2-2x+2y+5 given the constraint ( (x, y):x+y=4) Find the absolute maximum and minimum of f(x,y)=x2+y2-2x+2y+5 given the constraint [x,D): x<+y=4] Define g(x,y)=x2+y2-4 then g(x,y)=0 means x2+y2=4 Objective function f(x,y)=x2+y2-2x+2y+5 gives the surface f(x,y)=x2+y2-2x+2y+5 Constraint Curve limits the (x, y) values to only those points that satisfy g(x,)=0 where g(x,y)=x+y-4 Level Curve g(x, y)=0 f(x,y)=x2+y2-2x+2y+5 where g(x, y)=x+y2-4 9 The gradient V g is orthogonal to the level curve g(x,y)=0. Level Curve g (x,y)=0 f(x,y)=x2+y2-2x+2y+5 where g(, y)=x2+y2-4 IaEVG The gradient V g is orthogonal to the level curve g(x, y)=0. The gradient Vf is orthogonal to the level curve f(x, y)=Zo f(x,y)=x2+y2-2x+2y+5 Level Curve g(x,y)=0 where g(x,y)=x+y-4 vf 冒(的些P If f has a max or min at the point(a, b) on the constraint curve then Vf(a, b) points in the same direction as Vg(a, b)

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