Question

Suppose that f(x,y)=x^2+y^2+kxy. Find and classify the critical points and discuss how they change when k takes on different values.

Answer 1

I know that f_x=2x+ky and that f_y=2y+kx. I also know that critical points occur when both partial derivatives are zero... Also that each point is a maximum if both second differentials are negative, a minimum if both are positive, and a saddle point if they both have a different sign...... But the constant K is throwing me off... Any help would be greatly appreciated.

Answer 2

all you need to do really is solve the system
\(f_y=2x+ky=0\)
\(f_x=2y+kx=0\)

you can use substitutions, and you should get answers in terms of k
\(x=\{a_1,a_2\}\), \(y=\{b_1,b_2\}\)

the key is finding
\[D=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2\](you will find that the points a and b don't even matter here as the second derivatives are all constants)
for which
when \(D>0\) and \(f_{xx}>0\) is a relative min
when \(D>0\) and \(f_{xx}<0\) is a relative max
when \(D<0\) is a saddle point
and when \(D=0\) nothing can be determined with this method

Answer 3

I still don't understand.. substituting will still not get rid of the variable because I don't know what K is.

Answer 4

like I said, your critical points will be in terms of k
let k stand as a variable, just try to eliminate x and y

Answer 5

So, I got f_x=2x+ky
f_y=2y+kx.
I then did this:
f_x=2x+ky=0
f_y=2y+kx=0
I solved for y and substituted y in the f_x equation with the value I got. I then had:
y=kx/2
2x+ky=0------>2x+k(kx/2)
I then proceeded to solve for x and got...
x(4+k^2)=0
So...x=0?
I then plugged 0 into the y= equation
y=k(0)/2
and got y=0.
Hence (0,0) is a critical point, yes?
What's my next step?