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?
$\displaystyle u=e^{-x^2-y^2}(2x^2+3y^2)=2x^2e^{-x^2-y^2}+3y^2e^{-x^2-y^2}$
So $\displaystyle \frac{\partial{u}}{\partial{x}}=4xe^{-x^2-y^2}-4x^3e^{-x^2-y^2}-6xy^2e^{-x^2-y^2}$
and
$\displaystyle \frac{\partial{u}}{\partial{y}}=-4yx^2e^{-x^2-y^2}+6ye^{-x^2-y^2}-6y^3e^{-x^2-y^2}$
Ok, so factoring a little we see that
$\displaystyle u_x=e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)$
and $\displaystyle u_y=e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)$
So we need to find all $\displaystyle (x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0$
I get $\displaystyle (-1,0),(0,-1),(0,0),(0,1),(1,0)$
Now test them all in the discriminant.
$\displaystyle D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2$
And then draw the correct conclusions.
$\displaystyle u_x=0\Rightarrow{e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)}=0$
and
$\displaystyle u_y=0\Rightarrow{e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)}=0$
So we need to simultaneously solve these. But we know the exponential function is never zero. so we just need to solve
$\displaystyle 4x-4x^3-6xy^2=0$
and
$\displaystyle -4xy^2+6y-6y^3=0$
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:
$\displaystyle 2x^2 + 3y^2 = 2$
$\displaystyle 2x^2 + 3y^2 = 3$
This is the problem, what do I do from here?
10x again.
ok.. so your system of equations is:
$\displaystyle \frac{\partial u}{\partial x} = -2xe^{-x^2-y^2}(2x^2+3y^2-2)=0$--------(1)
$\displaystyle \frac{\partial u}{\partial y} = -2ye^{-x^2-y^2}(2x^2+3y^2-3)=0$--------(2)
if you set $\displaystyle x=0$ in (1), you should substitute that value in (2) to get values of $\displaystyle y$ corresponding to $\displaystyle x=0$. from here, you will get three values of $\displaystyle y$ corresponding to $\displaystyle x=0$.
if in (1), $\displaystyle x\not=0$, then it must be
$\displaystyle 2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2$..
if you substitute this in (2), you will get
$\displaystyle -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, $\displaystyle y=0$.
substitute this back to
$\displaystyle 2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2$
to get the specific value of $\displaystyle x$ which is not equal to 0 (corresponding to $\displaystyle y=0$).
you should get the values what Mathstud28 got..