Use the Lagrange method to find the points in $\mathbb{R}^3$ closest to the origins, and which are on the cone $z^2 = x^2 + y^2$ and also on the plane $x+2y=6$.
We want to minimize the distance from the origin to the point(s) $P$, thus we want to minimize $\sqrt{x^2+y^2+z^2}$, which is equal to minimizing the function $f(x,y,z)=x^2+y^2+z^2$. We minimize this subject to the constraints $g(x,y,z) = x^2+y^2-z^2$ and $h(x,y,z) = x+2y-6$.
Using the Lagrange method, we have the following set of equations:
$$\nabla f(x,y,z) = \lambda \nabla g(x,y,z) + \mu \nabla h(x,y,z) $$
$$\langle 2x,2y,2z \rangle = \lambda \langle 2x,2y,-2z \rangle + \mu \langle 1,2,0 \rangle $$
Solving the system of equations arising from this yields us the points $P = \big(\dfrac{6}{5}, \dfrac{12}{5}, \pm \dfrac{6}{\sqrt{5}})$
I'm wondering if my application of the Lagrange method here is correct.
Best Answer
The use of two Lagrangian multipliers to solve the problem seems to be entirely sound, and so I have really nothing to add there. But it's worth noting that you can actually get away with just one such multiplier!
Since $x^2+y^2=z^2$, we have $f(x,y,z)=x^2+y^2+z^2=2x^2+2y^2$. Then both this objective function and the constraint $g(x,y)=x+2y=6$ depend only $(x,y)$; all that remains is $x^2+y^2=z^2$, which doesn't constrain $x,y$ unless $z$ is known. So we can set aside this equation, and what we have is an optimization problem in $(x,y)$. Introducing a Lagrangian multiplier $\lambda$, we demand that $\nabla f = (4x,4y)=\lambda \nabla g = (\lambda,2\lambda)$. This implies $\lambda = 2y=4x$ and so $(x,y)=(\frac65,\frac{12}5).$ Then $z=\pm \sqrt{x^2+y^2}=\pm\frac{6}{\sqrt{5}}$ and the minimum distance is $\frac{12}{\sqrt{5}}.$