Some difficulties using Lagrange Multipliers

Question

I had to solve the following problem: Find the distance from the point $A(1,1,0)$ to the surface $xy+xz+yz=0$. Using Lagrange Multipliers I obtained the following system

$2(x-1)=t(y+z), 2(y-1)=t(x+z), 2z=t(y+x), xy+xz+yz=0$ and the distance is $2^{(1/2)}/3$

I had to use the software Mathematica to solve this system because I could't solve it by myself. But I am not satisfied, I would like to know a way to solve it without using Mathematica. I appreciate any help. Thanks.

Answer 1

Not too methodic, but... Sum the first 3 of your 4 equations. You get $$ 2(x+y+z)-4 = 2t(x+y+z) \quad \Rightarrow\quad x+y+z = \displaystyle\frac{2}{1-t}. $$ Now these equations become $2(x-1)=t\Big(\displaystyle\frac{2}{1-t}-x\Big)$, etc.; solving them you get $$ x=y=\frac{2}{(1-t)(2+t)}, z = \frac{2t}{(1-t)(2+t)}; $$ plugging it into $xy+xz+yz=0$ reveals $t=-1/2$.

So $x=y=8/9$, $z=-4/9$, and the required distance is $$ \sqrt{(x-1)^2+(y-1)^2+z^2}=\sqrt{\frac{1}{81}+\frac{1}{81}+\frac{16}{81}}=\frac{\sqrt{2}}{3} $$ as expected.

Answer 2

If I understood correctly, you wish to minimise

$$ (x-1)^2 + (y-1)^2 + z^2 $$

subject to the constraint $xy+xz+yz=0$.

One approach is to write down the Lagrangian function

$$ (x-1)^2 + (y-1)^2 + z^2 - \lambda(xy+xz+yz), $$

where $\lambda$ is the Lagrange multiplier for the constraint. We first calculate the first-order conditions (FOCs)

\begin{align*} 2(x-1)-\lambda (y+z) &= 0, \\ 2(y-1) - \lambda(x+z) &= 0, \\ 2z - \lambda (x+y) &= 0. \end{align*}

The FOCs, along with the constraint, give a system of four equations in four unknowns ($x,y,z,\lambda$).

The FOCs determine a linear system in $(x,y,z)$, so it seems fruitful to start from there:

\begin{align*} 2x -\lambda y -\lambda z &= 2, \\ - \lambda x + 2y - \lambda z &= 2, \\ - \lambda x - \lambda y + 2z &= 0. \end{align*}

This system gives \begin{align*} x &= -\frac{2}{(\lambda-1)(\lambda+2)}, \\ y &= -\frac{2}{(\lambda-1)(\lambda+2)}, \\ z &= -\frac{2\lambda}{(\lambda-1)(\lambda+2)}. \end{align*}

Crucially, this gives us that $x=y$ and $z=\lambda x = \lambda y$. Substituting this into the constraint yields

$$ (1+2 \lambda)x^2 = 0. $$

This gives us that $\lambda = -\frac{1}{2}$ (we can rule out $x=0$ given the expression we derived earlier), which we can plug into our solution for $(x,y,z)$ to find

\begin{align*} x &= \frac{8}{9}, \\ y &= \frac{8}{9}, \\ z &= -\frac{4}{9}. \end{align*}

Of course, one then needs to verify that the first-order conditions give a minimum, but I'll leave that verification to you.