In my economics class, we usually consider very simple optimization problems.
$$L(x,y,\lambda) = xy + \lambda (m-px-qy)$$ $$L_x = y - \lambda p = 0$$ $$L_y = x - \lambda q = 0$$
$$ \frac x q = \lambda = \frac y p$$
$$x = \frac {qy} p$$
$$m = p \frac{qy}p + qy = 2qy$$
$$y = \frac{m}{2q}$$
But suppose I have $$L(\Bbb x, \lambda ) = f(\Bbb x) + \lambda (m - \Bbb p \cdot \Bbb x)$$
And then I have
$$\nabla L =0$$ $$\nabla f(\Bbb x) = \lambda \nabla (\Bbb p \cdot \Bbb x)$$ $$m = f(\Bbb x)$$
How do I solve this for the $n$ dimensional case? Surely I can't do it like I did the two dimensional one where I equated $\lambda$
In the two dimension case, the gradient of $f$ was composed of two equations, one partial derivative wrt $x$ and one wrt $y$. Then $\bigtriangledown L =0$ is in fact a ystem of three equations with three unkonwns $x$, $y$, and $\lambda $.
In the $n$ dimensional case, you would have $n$ equations from $\bigtriangledown f (x)=0$ and one equation from $ \frac{dL}{ d \lambda}=0$. So infact you have $n+1$ equations with $n+1$ unkowns.
Usually in such problems first component $x_i$ is written in term of $\lambda$. Then all are sabstituted in the las equation $ \frac{dL}{ d \lambda}=0$ to get $\lambda$, lastly by substitution the components are recovered.