Local Minimum and Maximum; Saddle Point

**lilaziz1** · October 24th 2010, 06:13 PM

Hi everyone. I have a question on one of my homework problems.

Q. Find the local max./min. values and the saddle point(s) of the function $f(x,y) = (x^2+y^2)e^{y^2-x^2}$

So I started doing this problem and it was looking really messy so I went on cramster and got this:

Step 1

Step 2

Step 3

so critical points are (0,0), (1,0), and (-1,0)

Step 4

Step 5

at (0,0):

at (1,0):

at (-1,0):

Therefore, f has a local minimum of 0 at (0,0) and saddle points at (1,0) and (-1,0).

The answer is correct but the work makes no sense. When computing $f_x$ , $y^2e^{y^2-x^2}$ was completely neglected. $f_x$ should have been:

$f_x = 2xe^{y^2-x^2}-2x^3e^{y^2-x^2} -2xy^2e^{y^2-x^2} = 0$

Same problem with $f_y$

Is there something wrong with the solution or am I doing something wrong?

Thanks in advance!

**tonio** · October 25th 2010, 04:16 AM

Originally Posted by lilaziz1

Hi everyone. I have a question on one of my homework problems.

Q. Find the local max./min. values and the saddle point(s) of the function $f(x,y) = (x^2+y^2)e^{y^2-x^2}$

So I started doing this problem and it was looking really messy so I went on cramster and got this:

Step 1

Step 2

There's already a mistake here since you forgot to differentiate the second summand above. It should be:

$f_x=2xe^{y^2-x^2}-2x^3e^{y^2-x^2}-2xy^2e^{y^2-x^2}=0\Longrightarrow 2xe^{y^2-x^2}(1-x^2-y^2)=0$ , and this

alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative...

Likewise $f_y$ is wrong...you seem to have some problems with partial derivatives. Check this.

Tonio

Step 3

so critical points are (0,0), (1,0), and (-1,0)

Step 4

Step 5

at (0,0):

at (1,0):

at (-1,0):

Therefore, f has a local minimum of 0 at (0,0) and saddle points at (1,0) and (-1,0).

The answer is correct but the work makes no sense. When computing $f_x$ , $y^2e^{y^2-x^2}$ was completely neglected. $f_x$ should have been:

$f_x = 2xe^{y^2-x^2}-2x^3e^{y^2-x^2} -2xy^2e^{y^2-x^2} = 0$

Same problem with $f_y$

Is there something wrong with the solution or am I doing something wrong?

Thanks in advance!

.

**lilaziz1** · October 25th 2010, 08:05 AM

Ah okay. As I stated at the end of the thread, those mistakes are not mine. they are from cramster. I also corrected $f_x$ in the first thread. Anyhow, my question is what do you mean by, "this alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative..." How does it tell you that?

**tonio** · October 25th 2010, 09:30 AM

Originally Posted by lilaziz1

Ah okay. As I stated at the end of the thread, those mistakes are not mine. they are from cramster. I also corrected $f_x$ in the first thread. Anyhow, my question is what do you mean by, "this alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative..." How does it tell you that?

Check the corrected $f_x$ . This can be equal to zero if $1-x^2-y^2=0\Longleftrightarrow x^2+y^2=1$ , and this last eq.

describes the unit circle centered at the origin...

Tonio

**lilaziz1** · October 25th 2010, 12:31 PM

Yes I got that but how does this tell you that all the derivative are zero around the cirlce?

**tonio** · October 25th 2010, 08:37 PM

Originally Posted by lilaziz1

Yes I got that but how does this tell you that all the derivative are zero around the cirlce?

I already answered: because if a point $(x,y)$ belongs to THAT circle then inputting it in $f_x$ we get zero...what else?!

Tonio

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