I am working through some old test papers in preparation for exams an am trying to scout out potential sneaky questions that might be asked. I've stumbled across this one. Would you please verify or correct my solution. Thank you very much.
$\textbf{Question}$:
Consider the problem of finding the extrema of $f(x,y)$ with constraints $h_1(x,y) = 0$ and $h_2(x,y)=0$. All three $f,h_1,h_2$ are $C^2$ functions (I imagine they are trying to say second-order conditions should old). Suppose $(x_0,y_0)$ satisfies $h_1(x_0,y_0)=h_2(x_0,y_0)=0$ and $$ \nabla f(x_0,y_0) + \mu_1\nabla h_1(x_0,y_0) + \mu_2\nabla h_2(x_0,y_0) =(0,0) $$ Is there a unique $\mu_1,\mu_2 \in \mathbb{R}$ that a unique solution for the above exists while satisfying for some $\alpha \in \mathbb{R}$ $$ \nabla h_1(x_0,y_0) = \alpha \nabla h_2(x_0,y_0) $$
$\textbf{Solution}$
So what they gave us was the first-order neccessary conditions and essentially asks the Lagrangian to be zero. When this happens, $f$ is noted to be a multiple of the constraints. This would mean it is linearly dependent on it. I state this because I am trying to say that in finding a unique solution we need not bother ourselves with it. We shift our focus to $$ \mu_1\nabla h_1(x_0,y_0) + \mu_2\nabla h_2(x_0,y_0) $$ Which we can rewrite as $$ \mu_1\,\alpha \left[ \begin{matrix} (h_2)_x\\(h_2)_y \end{matrix}\right] + \mu_2 \left[ \begin{matrix} (h_2)_x\\(h_2)_y \end{matrix}\right] $$ Or for even further convenience $$\left[ \begin{matrix} \mu_1 \alpha & \mu_2 \\ \mu_1 \alpha & \mu_2 \end{matrix} \right] \left[ \begin{matrix} (h_1)_x \\ (h_2)_y \end{matrix} \right] \iff \mathbf{Ax} $$ We already note that the rows are factors of one another such that they are not linearly independent. Thus a unique solution does not exist. To be fancy, we can also test for non-zero determinant. $$ \det(\mathbf{A}) = \alpha\mu_1\mu_2 -\alpha\mu_1\mu_2= 0 $$ Thus we have linear dependance and non-unique solutions.
$\textbf{Extra:}$ (dont't have to answer if you want to)
They go on to ask for linearly independent $h_1$ and $h_2$ which satisfy at $(x_0,y_0)$ $$ \nabla h_1(x_0,y_0) = \alpha \nabla h_2(x_0,y_0) $$
I said that at (0,0) with $\alpha = 2$ and constraints: $$ h_1 = x^2 - y^2 $$ $$ h_2 = e^x - e^y$$ It would hold. Do you agree with this example.
Thank you for your time.