Usbrth

The question is simple: "find the point closest to the origin that is on the intersection line of $y+2z=12$ and $x+y=6$."

Normal method would use two Lagrange multipliers and get the point $(2,4,4)$. I get that. BUT I used the following method and failed to get the same result:

I designed an equivalent constraint from the 2 constraints given above:

$g(x,y,z) = (y+2z-12)^2 + (x+y-6)^2 = 0$

And then I tried to use only one Lagrange multiplier since we only have one constraint now. However, I cannot seem to get $(2,4,4)$ So my question is: what went wrong? Did I miss something?

$mathbfEDIT$: I previously posted a "solution" agreeing that the OP's method works, but I was incorrect; this is a correct (to the best of my knowledge) revision.

The method of Lagrange multipliers for a single constraint seeks to extremize a real-valued function $f(vecx)$ subject to the constraint $g(vecx)=0$. That is, the solution $vecx_0$ must extremize the restriction of $f$ to $S equiv g(vecx)=0 $. The method makes use of the fact that, if $nabla g(vecx_0) neq 0$, then $S$ is a differentiable surface in a neighborhood of $x_0$ with normal vector $nabla g(vecx_0)$, and since the restriction $f|_S$ is extremized at $x_0$, we must then have that $nabla f(vecx_0)$ is normal to the surface (or else $f$ would be increasing along a curve through $S$ with tangent at $vecx_0$ equal to the projection of $nabla f(vecx_0)$ onto the tangent space $T_vecx_0S$). That is, $nabla f(vecx_0) = lambda_0 nabla g(vecx_0)$ for some $lambda_0 in mathbbR$. So, if one defines the function
$$L(vecx,lambda) equiv f(vecx)-lambda g(vecx) $$
Then $nabla L(vecx_0,lambda_0)=0$ ($nabla$ here is the gradient in $n+1$ dimensions). The above logic and therefore this conclusion, however, were contingent on the statement that $nabla g(vecx_0) neq 0$. If this is not satisfied, the extremizer $vecx_0$ of $f|_S$ need not satisfy $nabla L(vecx_0,lambda) = 0$ for any $lambda in mathbbR$, and we may not be able to recover the solution $vecx_0$ by finding stationary points of $L$, and this is the precisely the problem in your approach.

Indeed, if $g$ is a sum of squares of functions, $g(vecx)=sum_k (g_k(vecx))^2$, then $g(vecx)=0 iff g_k(vecx)=0$ for each $k$, so for every $vecx_0 in S$
$$nabla g(vecx_0) = sum_k 2g_k(vecx_0) nabla g_k(vecx_0) = 0$$
So no point in $S$ satisfies the hypotheses necessary to rely on the Lagrange method.

A main reason is that, for the method of Lagrange multipliers, setting :

$$g(x,y,z) = (y+2z-12)^2 + (x+y-6)^2 colorred= 0tag1$$

(I understand that you want it to be zero for the equivalence with your two linear equations)

or setting

$$g(x,y,z) = (y+2z-12)^2 + (x+y-6)^2 colorred= ktag2$$

for any constant $k$ is the same. (constant $k$ disappears in differentiation).

Equation (2) describes a family of "russian dolls" cylinders $C_k$.

Edit 1 : in fact, encountering cylinders could be a good thing, because your problem could have been turned into this one ; take increasing radii circular cylinders $Gamma_R$ with common axis the intersection line $(L)$ of $y+2z=12$ and $x+y=6$ and radius $R$. Stop when $R$ is such that $Gamma_R$ passes through the origin. And this $R$ is the looked-for distance.

But the issue is that, cylinders $C_k$ described in the first part do not grow in the good way (in fact they are elliptical cylinders) : the value of $k$ such that $C_k$ passes through the origin cannot be related to a distance...

But, we could remedy to this situation by defining $L$ otherwise, as the intersection of planes

$$P_1=u_1x+v_1y+w_1z-h_1=0, textand P_2=u_2x+v_2y+w_2z-h_2=0,$$

• perpendicular one to the other ($u_1u_2+v_1v_2+w_1w_2=0$),




• with normalized coefficients ($u_k^2+v_k^2+w_k^2=1$, $k=1,2$).

Then, by replacing $(y+2z−12)^2+(x+y−6)^2$ by $P_1^2+P_2^2$ we have now an expression which is the square of the distance to axis $(L)$, and we can use what we have said before.

Edit 2 : in fact one cannot rigorouly speak of $vecgrad(g)$ on $g(x,y,z)=0$ because this expression is equivalent to the equation of the straight line $(L)$ and a straight line in 3D has no gradient.