图解拉格朗日乘子法

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)

...展开详情
LeisureZhao