I recently saw the following problem in calculus:
Let $f(x,y)$ and $g(x,y)$ be smooth functions on $\mathbb{R}^2$, $S=\{(x,y)|f(x,y)=0\}$ and $p=(a,b)\in S$. If $\frac{\partial f}{\partial x}(p)=-4$, $\frac{\partial f}{\partial y}(p)=2$, $\frac{\partial g}{\partial x}(p)=12$, $\frac{\partial g}{\partial y}(p)=-6$, and $$ \begin{bmatrix} \frac{\partial^2 f}{\partial x^2}(p) & \frac{\partial^2 f}{\partial x\partial y}(p) \\ \frac{\partial^2 f}{\partial y\partial x}(p) & \frac{\partial^2 f}{\partial y^2}(p) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ \end{bmatrix}$$ $$\begin{bmatrix} \frac{\partial^2 g}{\partial x^2}(p) & \frac{\partial^2 g}{\partial x\partial y}(p) \\ \frac{\partial^2 g}{\partial y\partial x}(p) & \frac{\partial^2 g}{\partial y^2}(p) \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -1 & 2 \\ \end{bmatrix}$$ Show that there is $R>0$ such that $g(p)<g(q)$ for all $q\in S \cap \{(x,y)|x^2+y^2<R^2\}$
I think this is related to Lagrange multiplier because $\nabla f(p)=-3\cdot\nabla g(p)$ and the statement is about extrema of $g$ with restriction $S$. But I don't know how to examine if the local minimum of $g$ occurs on $p$ under the condition that second partial derivatives is known. I am not sure if this is related to the second partial derivative test which is always used to examining local extrema of a function without restriction.
Hope someone could help me with this one. Thanks!
By the implicit function theorem near $p$ in $S$ you can write $y=h(x)$. Then consider $p(x)=g(x,h(x))$. You need to compute $p'(a)$ and $p''(a)$. Since $f(x,h(x))=0$ you can differentiate twice to find $h'(a)$ and $h''(a)$. Using those and the chain rule you can find $p'(a)$ and $p''(a)$. Let me know if you need more details.