Lagrange Multipliers-higher dimensional I'm studying for an exam and am trying to work out this example. 
Use Lagrange Multipliers to find the maximum value of $(xv-yu)^2$ subject to the constraints $x^2+y^2=a^2$ and $u^2+v^2=b^2$. 
My guess how to start is to let $X=(x,y)$ and $Y=(u,v)$ so $f(X,Y)=(xv-yu)^2$. Then take $Df_X$ and $Df_Y$ and go from there. Any help would be appreciated.
 A: I don’t think you even need to take $X,Y$ as you have said. We can define
$g:\mathbb{R}^4\rightarrow \mathbb{R}^2$ as 
$$g(x,y,u,v)=(x^2+y^2-a^2,u^2+v^2-b^2)\equiv(g_1,g_2)$$ 
For the Lagrange Multipliers Theorem to apply, we need the preimage $g^{-1}(\{0\})$ to be a $2$-dimensional sub-manifold of $\mathbb{R}^2$. It is sufficient to show that
$$\text{rk}\left(\begin{array}{cc}
\nabla g_1\\
\nabla g_2 \end{array}\right)=2$$
We calculate $\nabla g_1=(2x,2y,0,0)$ and $\nabla g_2=(0,0,2u,2v)$, so that our matrix has rank 2 everywhere except when $x=y=0$ or $u=v=0$. But assuming $a,b$ are nonzero, this is not an issue, as none of these points lie in $g^{-1}(\{0\})$. So we can apply the Lagrange Multipliers Theorem.
So there exist $\lambda_1,\lambda_2$ scalars such that
$$\nabla f=\lambda_1\nabla g_1 + \lambda_2\nabla g_2$$
We calculate $\nabla f=(xv-yu)*(2v,-2u,-2y,2x)$, where the multiplication is component-wise. This gives us the system of equations
$$\begin{array}{cc}
(xv-yu)v&=\lambda_1 x\\
-(xv-yu)u&=\lambda_1 y\\
-(xv-yu)y&=\lambda_2 u\\
(xv-yu)x&=\lambda_2 v
\end{array}$$
which you can then solve to find critical points. My next step from here would be to eliminate $\lambda_1,\lambda_2$.
