I wanted to try this problem: $ax+by+cz=d$. Find the lowest possible value for $x+y+z$
So I wanted to try Lagrange multipliers.
Minimize $f(x,y,z) = x+y+z$ subject to constraint $g(x,y,z) = ax + by + cz = d$.
Then $\triangledown f = \langle 1, 1, 1 \rangle$ and $\triangledown g = \langle a, b, c \rangle$. The equations to be solved are:
$1 = a \lambda $
$1 = b \lambda $
$1 = c \lambda $
$ax + by + cz = d$
Then $a = b = c = \dfrac{1}{\lambda}$ and
$\dfrac{x}{\lambda} + \dfrac{y}{\lambda} + \dfrac{z}{\lambda} = d$, or $\lambda = \dfrac{x+y+z}{d}$
But where do I go from here? Aren't I supposed to get $\lambda$ in terms of $a,b,c,d$? How can I do that when the $x,y,z$ terms disappear as a result of taking the derivatives of $f$ and $g$?
The method is working fine. If $a=b=c$ ($=\lambda\neq 0$, say) then your constraint is $$\lambda(x+y+z)=d\implies x+y+z=\frac d{\lambda}$$ So every point on the constraint surface is a minimum (and a maximum).
If you don't have $a=b=c$ then the method correctly shows that a minimum does not exist. For example, suppose the constraint is $$x-y+z=0$$ then $(x,y,z)=(N,N,0)$ is a solution for any $N$ and we can make $x+y+z=2N$ as small (or as large) as we desire.