Deducing Lagrange multipliers How much do I need to know to be able to deduce the method of Lagrange multipliers/prove why it works? 
I have had real analysis. Is there a way to understand why it works without having to learn differential forms?
Of course, I want to be able to do it rigorously.
 A: I will sketch the proof of the "Necessary conditions of first order in constrained optimization." I will write $\approx$ to mean that the spaces are canonically isomorphic. The space of continuous linear functions $V \to W$ will be denoted by $\mathscr{L}_W(V).$ Notice that if $T \in \mathscr{L}_{\Bbb R}(\Bbb R^q),$ then $T$ is a $1 \times q$ matrix, so if $S$ is a $q \times q$ matrix, then $TS$ is a $1 \times q$ matrix. The following sketch can be greatly generalized up to normed and complete vector spaces (and if my guess is right, up to differential varieties).
Lagrange multipliers. Let $A$ be an open subset of $\Bbb R^p \times \Bbb R^q \approx \Bbb R^{p + q}$ and $f, g$ two functions from $A$ to $\Bbb R$ and $\Bbb R^q,$ respectively. Assume $f$ is differentiable and $g$ is of class $\mathscr{C}^1_{\Bbb R^q}(A).$ Suppose that $(x^*,y^*) \in g^{-1}(\{0\})$ is an optimizer of the problem $$\arg \min f(x,y) \quad \mathrm{s.t.} \quad g(x,y) = 0.$$
If $\mathbf{D}_2g(x^*,y^*) = \dfrac{\partial g}{\partial y}\Bigg|_{(x^*,y^*)}$ is a linear homeomorphism (invertible and continuous, with a continous inverse), then there exists a $\lambda \in \mathscr{L}_{\Bbb R}(\Bbb R^q) \approx \Bbb R^q$ such that $\mathbf{D}f(x^*, y^*) = \lambda \mathbf{D}g(x^*, y^*).$
Sketch. Use implicit function theorem to get a $\mathscr{C}^1$ function $h$ from some open set in $\Bbb R^p$ to $\Bbb R^q$ such the relation $g(x,y) = 0$ is the same as $h(x) = y$ and the equality $h(x^*) = y^*$ holds. Define $\dot f(x) = f(x, h(x)).$ Then $\dot f$ has an optimum in $x^*$ and since it is the composition of two differentiable functions (though $h$ is $\mathscr{C}^1,$ $f$ is only assumed differentiable), it follows that $\mathbf{D} \dot f(x^*) = 0.$ However, $$\mathbf{D} \dot f(x) = \dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} \mathbf{D}h(x).$$
The implicit function theorem gives $\mathbf{D}h(x) = -\left[\dfrac{\partial g}{\partial y}\right]^{-1} \dfrac{\partial g}{\partial x}$; hence $$\mathbf{D} \dot f(x) = \dfrac{\partial f}{\partial x} - \dfrac{\partial f}{\partial y} \left[\dfrac{\partial g}{\partial y}\right]^{-1} \dfrac{\partial g}{\partial x} = \dfrac{\partial f}{\partial x} - \lambda \dfrac{\partial g}{\partial x},$$
with $\lambda = \dfrac{\partial f}{\partial y} \left[\dfrac{\partial g}{\partial y}\right]^{-1} \in \mathscr{L}_{\Bbb R}(\Bbb R^q) \approx \Bbb R^q.$ This proved $$\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}$$ and obviously $$\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}.$$ Thus, $\mathbf{D}f(x^*, y^*) = \lambda \mathbf{D} g(x^*, y^*).$ C.Q.F.D.
A: The basic idea is that at the critical points of the constrained function, its gradient must be parallel to the gradient of the constraint, which is given as a level curve of some other function. Equivalently, the tangents vanish simultaneously at these points. Showing this in detail doesn’t require a knowledge of differential forms; standard multivariable calculus suffices.
