Finding extreme values in constraints using Lagrange multipliers
This question may sound repetitive, but please give it a read. I read about Lagrange multipliers, and I know the formula. However, I was trying to get the intuition behind this. I read the posts How do Lagrange multipliers work to find the lowest value of a function subject to a constraint? and .
There is just one point which I'm unable to digest. Any explanations on it would be very helpful. What the above two posts say is : Therefore, if π(π,π) was a maximum (or a minimum) of π on π(π₯,π¦)=π, gradient vector of π at (π,π) would be have dot product zero with every vector in the line tangent to π at (π,π). But that means that the gradient vector of π is orthogonal to the tangent line of π, which means that the gradient of π must be parallel to (i.e. a scalar multiple of) the gradient of π.
If vector of f is orthogonal to the tangent line of g, how does it imply that gradient of f must by parallel to gradient of g. In 2-D space, it makes sense. But what if say "vector of f is along x", "tangent of g is along y", and "gradient of f is along z, but gradient of g is along x?". Is this case not possible? If not, why?
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$\begingroup$Usually, if $g:\mathbb R^n \to \mathbb R$ is a $C^1$ fct, then for a given $c\in R(g)$ (is a proper value,) the set $g^{-1}(c)$ is a closed manifold of dimension $n-1$. This means that the tangent space at any of it's points is one codimensional, hence it's orthocomplement is 1 dimensional and is completely described by a vector, i.e. the normal vector of the tangent space. Since for any $x\in g^{-1}(c)$ $g(x)=c$, $\mathrm{grad} \ g(x)$ has to be orthogonal to the tangent space at $x$. Now if the function $f$ has an extremum at point $x\in g^{-1}(c)$, this means that $\mathrm{grad }\ f \cdot t $ has to be zero for any tangent vector $t$ at $x$, hence it is orthogonal to the tangent space at $x$, and so it is parallel to $\mathrm{grad }\ g(x)$.
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