Let $M_m$ be a $C^1$ surface in $R^n$, and let $y$ be a point in $R^n$ not on $M_m$. If $x$ is a point on $M_m$ that minimizes or maximizes the distance to $y$, how would one prove that the line joining $x$ and $y$ is perpendicular to the surface at $x$ (ie, perpendicular to the tangent space at $x$)?
(If it helps at all, this is on the sections about Lagrange multipliers and the second derivative test.)
I think that it's the following.
Define $f: M_m \to \mathbb{R}$ as $f(p)=|p-y|^2$, where $|\cdot |$ is the usual Euclidean norm on $\mathbb{R}^n$. Then $f$ is differentiable on $M_m$ because it is the restriction of a differentiable function on $\mathbb{R}^n$. Notice that for any $p \in M$ and $v \in T_pM$, the differential $df_p: T_pM \to \mathbb{R}$ is given by $df_p(v) = 2(p-y)\cdot v$, where $\cdot$ is the usual dot product in $\mathbb{R}^n$. The reason is that if $\gamma: (-\epsilon,\epsilon) \to M_m$ is any smooth curve with $\gamma(0)=p$, then $$df_p(\gamma'(0)):=(f \circ \gamma)'(0) = \frac{d}{dt}|\gamma(t)-y|^2\bigg|_{t=0} = 2(p-y)\cdot \gamma'(0)$$
In order to show that the line connecting $x$ and $y$ is perpendicular to $T_xM$, we need to show that $(y-x)\cdot v=0$ for all $v \in T_xM$. In order to prove this, notice that $df_x=0$ because $f$ achieves a global minimum (or maximum) at $x$, and thus for any $v \in T_xM$ we have that $2(y-x)\cdot v = df_x(v)=0$, which shows that $(y-x)\cdot v=0$.