Question about Lagrange multipliers and combining constraints

Question

Hi I have this question about Lagrange multipliers and specifically when there are 2 constraints given. The standard answer to this question uses the lagrangian and 2 constraints with 2 extra variables (lambda 1 and lambda 2). I understand this method completely fine, but when I first attempted this problem, I tried combining the 2 constraints given into one constraint (i.e. putting x=2z+3 into the other constraint) and then using Lagrange multipliers with only 1 constraint. However, this gave a completely different result! (The standard solution was sqrt(2) but I got 1 with my method). Can anyone explain why these 2 methods give different results?

Answer 1

Notice that replacing two constraints with one is not valid. Just to make this as straightforward as possible, consider the system of linear equations $x+y=0$ and $x-y=0$. The only solution, of course, is $x=y=0$. However, if we add the equations, eliminating $y$, we obtain $2x=0$, which is an entire line. Obviously, this alone is not enough information.

When you combined your two constraints, you went from a curve of eligible points to a surface of eligible points (a cylinder over an ellipse in the $yz$-plane). This allows too many eligible points.

Answer 2

There must be an error in your implementation of the 2nd method (since the ellipse you get by combining the two constraints does not have any point at distance 1 from the origin, whereas you claim to have found one!)

It is not completely clear from your question what you did. Going to only one constraint would logically require reducing the problem to two coordinates only, and from what you write you seem to have eliminated $x$, which leaves you with the equation in the $yz$-plane: $$ y^2+4(z-3/2)^2=1 $$ Still the function to be minimized remains the distance to the origin in the original problem, so in 3 dimensions! If you miss that, than indeed the point $$(y,z)=(0,-1)$$ in your restricted space gives you distance 1, so maybe that's what you have done. But the distance should be computed by (implicitly) including the $x$-coordinate again, which will extend the point to $(2z+3,y,z)$ so the quadratic form to minimize now isn't $y^2+z^2$ but: $$y^2+5z^2+12z+9$$ This (accidentally?) does not change the location where you find the minimum but its value at $(y,z)=(0,-1)$ will now of course nicely be $\sqrt{2}$ again.