Show that f has a stationary point at (0, 0) for every k ∈ R

[cathal84]

Homework Statement

Let f(x, y) = x^2 + kxy + y^2 , where k is some constant in R. i. Show that f has a stationary point at (0, 0) for every k ∈ R

Homework Equations

...

The Attempt at a Solution

I may have the solution or i may have gone completely wrong I am not entirely sure.
i first found the derivative of f(x,y) with respect to x it was 2x+ky
then found the derivative of f(x,y) with respect to y it was 2y+kx
i then let both of them equal 0

then i solved 2x+ky=0 looking for a value of x and i got x=-ky/2
i then put this value for x back into 2y+kx=0 initially looking for a value of y but then i got y to cancel and i got a value for k instead. i got k=2.

so then i rewrote my derivatives as 2x+2y=0
and 2y+2x=0
which i have just realized are the same equation. anyway,
i then went ahead and tried to solve one of them for a value of x so i could sub it back
and i got x=-y
then i went and subbed this new value for x back into the equation and I am getting 0=0
have i answered the equation at all?
would greatly appreciate any input.
if i am completely wrong it would help greatly if someone pointed me in the right direction
thanks

 

[Ray Vickson]

Homework Statement

Let f(x, y) = x^2 + kxy + y^2 , where k is some constant in R. i. Show that f has a stationary point at (0, 0) for every k ∈ R

Homework Equations

...

The Attempt at a Solution

I may have the solution or i may have gone completely wrong I am not entirely sure.
i first found the derivative of f(x,y) with respect to x it was 2x+ky
then found the derivative of f(x,y) with respect to y it was 2y+kx
i then let both of them equal 0

then i solved 2x+ky=0 looking for a value of x and i got x=-ky/2
i then put this value for x back into 2y+kx=0 initially looking for a value of y but then i got y to cancel and i got a value for k instead. i got k=2.

so then i rewrote my derivatives as 2x+2y=0
and 2y+2x=0
which i have just realized are the same equation. anyway,
i then went ahead and tried to solve one of them for a value of x so i could sub it back
and i got x=-y
then i went and subbed this new value for x back into the equation and I am getting 0=0
have i answered the equation at all?
would greatly appreciate any input.
if i am completely wrong it would help greatly if someone pointed me in the right direction
thanks

You have not dealt with the question asked: it asked you to show that $(x,y) = (0,0)$ is a stationary point, no matter what the value of $k$ may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.

 

[cathal84]

You have not dealt with the question asked: it asked you to show that $(x,y) = (0,0)$ is a stationary point, no matter what the value of $k$ may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.

thanks for your reply,

You have not dealt with the question asked: it asked you to show that $(x,y) = (0,0)$ is a stationary point, no matter what the value of $k$ may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.

thank you this has answered my question