1. ## Partial derivatives

What are the Partial derivatives for this function?

I came up with these expressions:

2x^2 + 3y^2 = 2

2x^2 + 3y^2 = 3

It's not logical, right?

2. Originally Posted by asi123
What are the Partial derivatives for this function?

I came up with these expressions:

2x^2 + 3y^2 = 2

2x^2 + 3y^2 = 3

It's not logical, right?
$u=e^{-x^2-y^2}(2x^2+3y^2)=2x^2e^{-x^2-y^2}+3y^2e^{-x^2-y^2}$

So $\frac{\partial{u}}{\partial{x}}=4xe^{-x^2-y^2}-4x^3e^{-x^2-y^2}-6xy^2e^{-x^2-y^2}$

and

$\frac{\partial{u}}{\partial{y}}=-4yx^2e^{-x^2-y^2}+6ye^{-x^2-y^2}-6y^3e^{-x^2-y^2}$

3. Originally Posted by Mathstud28
$u=e^{-x^2-y^2}(2x^2+3y^2)=2x^2e^{-x^2-y^2}+3y^2e^{-x^2-y^2}$

So $\frac{\partial{u}}{\partial{x}}=4xe^{-x^2-y^2}-4x^3e^{-x^2-y^2}-6xy^2e^{-x^2-y^2}$

and

$\frac{\partial{u}}{\partial{y}}=-4yx^2e^{-x^2-y^2}+6ye^{-x^2-y^2}-6y^3e^{-x^2-y^2}$

Ok, now I need to find Maxima and minima, so I compare those to 0, right?

And then I come up with what I showed you...

4. Originally Posted by asi123
Ok, now I need to find Maxima and minima, so I compare those to 0, right?

And then I come up with what I showed you...
Ok, so factoring a little we see that

$u_x=e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)$
and $u_y=e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)$

So we need to find all $(x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0$

I get $(-1,0),(0,-1),(0,0),(0,1),(1,0)$

Now test them all in the discriminant.

$D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2$

And then draw the correct conclusions.

5. Originally Posted by Mathstud28
Ok, so factoring a little we see that

$u_x=e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)$
and $u_y=e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)$

So we need to find all $(x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0$

I get $(-1,0),(0,-1),(0,0),(0,1),(1,0)$

Now test them all in the discriminant.

$D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2$

And then draw the correct conclusions.

One second , you did $(x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0$
and came up with this equations:

2x^2 + 3y^2 = 2

2x^2 + 3y^2 = 3

right?

How did you solved that? how did you come up with $(-1,0),(0,-1),(0,0),(0,1),(1,0)$?

6. Originally Posted by asi123
One second , you did $(x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0$
and came up with this equations:

2x^2 + 3y^2 = 2

2x^2 + 3y^2 = 3

right?

How did you solved that? how did you come up with $(-1,0),(0,-1),(0,0),(0,1),(1,0)$?
$u_x=0\Rightarrow{e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)}=0$

and

$u_y=0\Rightarrow{e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)}=0$

So we need to simultaneously solve these. But we know the exponential function is never zero. so we just need to solve

$4x-4x^3-6xy^2=0$

and

$-4xy^2+6y-6y^3=0$

7. If this is a maxima/minima problem, have you tried using Lagrange Multipliers? I think that technique would make things a "little" easier. When I find the time later, I will solve the problem using this alternative way...

--Chris

8. isn't Lagrange Multipliers is over do for this question? I found the Partial derivatives, compared them to 0 and came up with 2 equation that I can't find solution to:

$2x^2 + 3y^2 = 2$

$2x^2 + 3y^2 = 3$

This is the problem, what do I do from here?

10x again.

9. Originally Posted by asi123
isn't Lagrange Multipliers is over do for this question? I found the Partial derivatives, compared them to 0 and came up with 2 equation that I can't find solution to:

$2x^2 + 3y^2 = 2$

$2x^2 + 3y^2 = 3$

This is the problem, what do I do from here?

10x again.
Where are you getting these equations?!

${2x^2+3y^2=2}\brace{2x^2+3y^2=3}$ does not have any solutions!

We can see this since if we see that the left hand sides of both are equivalent we can set them equatl to each other getting $2=3$

10. This is what I did (pic)

11. ok.. so your system of equations is:

$\frac{\partial u}{\partial x} = -2xe^{-x^2-y^2}(2x^2+3y^2-2)=0$--------(1)

$\frac{\partial u}{\partial y} = -2ye^{-x^2-y^2}(2x^2+3y^2-3)=0$--------(2)

if you set $x=0$ in (1), you should substitute that value in (2) to get values of $y$ corresponding to $x=0$. from here, you will get three values of $y$ corresponding to $x=0$.

if in (1), $x\not=0$, then it must be
$2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2$..
if you substitute this in (2), you will get
$-2ye^{-x^2-y^2}(\underbrace{2x^2+3y^2}_{=2}-3)=-2ye^{-x^2-y^2}(2-3)=0$
you will get from here, $y=0$.
substitute this back to
$2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2$
to get the specific value of $x$ which is not equal to 0 (corresponding to $y=0$).

you should get the values what Mathstud28 got..