Find the critical points of f ( x , y ) = x 2 y e − x − y . x = 0 and x = 2 both satisfy f x = 0 , f y = 0, and when x = 2 , y = 1. But when x = 0 , y, x = 0 , y is arbritary. So how can I determine whether the critical point at 0 is max/min/saddle?

Question

Find the critical points of   f  (  x  ,  y  )  =      x    2    y      e          −      x      −      y        .  x  =  0  and   x  =  2 both satisfy       f    x    =  0  ,      f    y    =  0, and when   x  =  2  ,  y  =  1. But when   x  =  0  ,  y,  x  =  0  ,  y is arbritary. So how can I determine whether the critical point at   0 is max/min/saddle?

Kelly Ibarra · Accepted Answer

The set of critical points is   {  (  0  ,  y  )  :     y  ∈      R    }  ∪  {  (  2  ,  1  )  }. The Hessian of   f is  H  f  (  x  ,  y  )  =      [                            y                      e                          −              x              −              y                                (                      x            2                    −          4          x          +          2          )                          x          (          2          −          x          )                      e                          −              x              −              y                                (          1          −          y          )                                      x          (          2          −          x          )                      e                          −              x              −              y                                (          1          −          y          )                                      x            2                                e                          −              x              −              y                                (          y          −          2          )                      ]  The critical point   (  2  ,  1  ) is clearly a local maximum, since   H  f  (  2  ,  1  ) has deteminant   8      e          −      6        &amp;gt;  0 and trace   −  6      e          −      3        &amp;lt;  0This is sufficient to deduce that the points   {  (  0  ,  y  )  :     y  &amp;gt;  0  } are all local minimums,   {  (  0  ,  y  )  :     y  &amp;gt;  0  } are local maximums, and   (  0  ,  0  ) is a saddle.If you wonder what a curve of critical points graphically means, think of a half pipe placed horizontally on the ground: the set of points that touch the ground is all made of local (global too...) minimums. For a more mathematical example, just consider the shape of the surface   f  (  x  ,  y  )  =      x    2  .

Find the critical points of f(x,y)=x^(2)ye^(−x−y). x=0 and x=2 both satisfy f_x=0,f_y=0, and when x=2,y=1. But when x=0,y is arbritary. So how can I determine whether the critical point at 0 is max/min/saddle?

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