Finding Local/Absolute Extrema of f(x,y)=x^2+y^2

  • Thread starter Thread starter chunkyman343
  • Start date Start date
  • Tags Tags
    Extrema
chunkyman343
Messages
4
Reaction score
0
]I have to find the local/absolute extrema for the following function:
f(x,y)=x^2+y^2

bounded by the triangle x=0,y=0,y+2x=2

So far i have:
fx(x,y)=2x
fy(x,y)=2y
fxx(x,y)=2
fyy(x,y)=2
fxy&fyx(x,y)=0

critical pts at (0,0,0)
domain: 0<=x<=1, 0<=y<=2

i don't know what i should do next?
 
Physics news on Phys.org
Since it's only a function of two dimensions, the critical point you get is actually at (0,0). Notice that there are precisely two points where a maximum or a minimum can occur: At a critical point, or on the boundary of the domain. So now you have to check the boundary for any maxima (that you found a critical point on the boundary is pure coincidence; the rest of the boundary still needs to be checked)
 
So this is what i get:

local max: (0,2,4)
abs. min:(1,0,1)
abs. max:(1,2,5)
 
What happened to the critical point you found earlier?
 
(0,0,0) is an absolute min.

how does that look now?
 
with (1,0,1) being a local min.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top