Calculus 3

1. Jul 23, 2009

chunkyman343

]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 dont know what i should do next?

2. Jul 23, 2009

Office_Shredder

Staff Emeritus
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)

3. Jul 23, 2009

chunkyman343

So this is what i get:

local max: (0,2,4)
abs. min:(1,0,1)
abs. max:(1,2,5)

4. Jul 24, 2009

Office_Shredder

Staff Emeritus
What happened to the critical point you found earlier?

5. Jul 24, 2009

chunkyman343

(0,0,0) is an absolute min.

how does that look now?

6. Jul 24, 2009

chunkyman343

with (1,0,1) being a local min.