From zeilberg@math.rutgers.edu  Mon Oct  9 18:06:25 2006
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Date: Mon, 9 Oct 2006 18:05:50 -0400 (EDT)
From: Doron Zeilberger <zeilberg@math.rutgers.edu>
Message-Id: <200610092205.k99M5om22473@math.rutgers.edu>
To: zeilberg@math.rutgers.edu
Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu
Subject: Today's "quiz"
Status: RO
Content-Length: 1491


Dear Students,

Most people got it almost right. Quite a few
people confused MINIMUM POINT (location) with
MINIMUM VALUES (elevation at the minimum point).

Please remember that JUST this coming Thurs. (Oct. 12) there
is no Tutoring at Hill 705.

  Good luck in the exam, and get a good night's sleep on Wed.
If you know how to do all the homework problems (correctly) then
you should do really well.

                 Best wishes

                         Dr. Z.
  
------Solutions to the quiz of Oct. 9, 2006--------------------

Question: Find the local maximum and minimum values and saddle
point(s) of the function

f(x,y)=1+4x-8y+2x^2+8y^2

Solution:

Step ONE: Find f_x, f_y, f_xx, f_xy, f_yy

f_x = 4+4x   f_y = -8+16y

f_xx = 4   f_xy = 0  f_yy = 16


Step TWO: Set BOTH f_x and f_y to zero and solve
the SYSTEM of two equations and two unknowns

4+4x=0    -8+16y=0

gives x=-1  y=1/2

So, there is only ONE critical point in this problem:
(-1,1/2)

Step 3: Plug this point (in other problems you have to do
it for each point) into f_xx, f_xy, f_yy and compute
D=f_xx f_yy -(f_xy)^2

D=4(16)-0^2=64

Since D>0 it is EITHER a max or a min (i.e. there is NO way it
is a saddle point).

Since f_xx>0 it is a LOCAL MINIMUM.

Local Minimum : (-1,1/2) 

Local Maximum : None

Saddle Points: None

Step 4: (IMPORTANT!)
To get the VALUES you plug the point(s) into f

Local Minimum Values : f(-1,1/2) =1-4-8(1/2)+2(-1)^2+8(1/2)^2=-3

Local Maximum  Values: None

-------------------