How to find the Minimum of the Function. Minimize $xy$ on the ellipse $b^2x^2+a^2y^2=a^2b^2$
So what I did first was take the gradient of $f$ and $g$ 
$∇ f = (y,x)$    $\qquad$            $∇ g= (2x^2b^2,a^22y)$ 
Then we do the lagrange multiplier 
$y= 2x^2b^2λ$
$x=2ya^2λ$
Then we equate the functions. By multiplying.
$(y= 2x^2b^2λ)y^2a^2$
$(x=2ya^2λ )x^2b^2$

$y^3=2x^2y^2a^2b^2λ$
$x^3=2x^2ya^2b^2λ$
But this is where I get stuck. Setting them equal to each other gets me this.
$2x^2y^2a^2b^2= 2x^2ya^2b^2$
$y=1$ 
I think there is another method to solve this problem than the one I specifed does anyone know what that method is?
 A: $$L(x,y,z)=xy+\lambda(b^2x^2+a^2y^2-a^2b^2)$$
$$\nabla_{x,y,\lambda} L(x,y,z)=\langle y+2\lambda b^2 x, x+2\lambda a^2 y , b^2x^2+a^2y^2-a^2b^2 \rangle=\langle 0,0,0 \rangle$$

$$y=-2 \lambda b^2 x$$
$$x=-2 \lambda a^2 y$$

assuming $\lambda \ne0$:
$$\frac{y}{x}=-2 \lambda b^2$$
$$\frac{y}{x}=\frac{1}{-2 \lambda a^2}$$

$$-2 \lambda b^2=\frac{1}{-2 \lambda a^2}$$
or 
$$\lambda=\frac{\pm1}{2 a b}$$
$$y=\mp \frac{ b}{a}x$$
$$b^2x^2+a^2(\frac{b^2}{a^2})x^2=a^2b^2$$
$$x=\pm\frac{a}{\sqrt2}$$
$$y=\mp\frac{b}{\sqrt2}$$

$$xy=\frac{-ab}{2}$$
A: HINT: i have the system
$$y+\lambda(2b^2x)=0$$
$$x+\lambda(2a^2y)=0$$
$$b^2x^2+a^2y^2-a^2b^2=0$$
A: I would solve this system instead by solving a much simpler version via transformation. No calculus necessary even!
It would be much easier if this problem were symmetric, but in its current form it's not. However, rewrite the constraint in standard ellipse form and the symmetry should jump out at you:
$$(\frac{x}{a})^2+(\frac{y}{b})^2=1$$
Something involving $x$ alone squared plus the square of something involving $y$ alone. Lets use that observation and try the substitution $u=\frac{x}{a}$, $v=\frac{y}{b}$.
The problem then becomes:
Minimize $uv$ subject to $u^2+v^2=1$ (we can drop positive constants from the objective function without changing anything).
Hopefully it's now apparent (or at least more readily seen upon reflection -- if not, perhaps see if you can construct a proof by contradiction) that $|u|=|v|$ at the minimum, and that they should have opposite signs to force their product negative, i.e. $u=-v$.
This only happens twice on the constraint, and the value of the objective will be the same both times, namely $-u^2$, since the quadratic eliminates the positive/negative distinction.
More importantly, the 45 degree line intersects the unit circle at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$, which, using our transformation to get back to $x$ and $y$, brings us the solution:
$$x=\frac{a}{\sqrt{2}}, y = \frac{b}{\sqrt{2}}$$
This is related to the famous Nash bargaining solution (would be basically equivalent if you were maximizing)
