Lagrange multipliers from hell

Question

I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:

"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".

What I did:

the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$

$C_2(x,y,z)=x-2z$

$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:

$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$

Let's derive and solve when derivatives are zero:

$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$

$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$

$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$

$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$

$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$

Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?

Answer 1

The circle can be parametrized as $$ x = 2 + \frac{4}{\sqrt 5} \sin \theta, \; y = 2 + 2 \cos \theta, \; z = 1 + \frac{2}{\sqrt 5} \sin \theta. $$

The squared distance of such a point from the origin is $$ f(\theta) = 13 + 4 \sqrt 5 \sin \theta + 8 \cos \theta. $$ Derivative is $$ f'(\theta ) = 4 \sqrt 5 \cos \theta - 8 \sin \theta. $$ So, the two extrema occur where $\tan \theta = \frac{\sqrt 5}{2};$ from $1 + \tan^2 \theta = \sec^2 \theta$ we get $$\sec^2 \theta = \frac{9}{4}, \cos^2 \theta = \frac{4}{9},$$ and $$ \cos \theta = \pm \frac{2}{3}, \; \sin \theta = \pm \frac{\sqrt 5}{3}, $$ with matching $\pm.$

The farthest point from the origin is $$ \left( \frac{10}{3}, \; \frac{10}{3}, \; \frac{5}{3} \right), $$ the nearest $$ \left( \frac{2}{3}, \; \frac{2}{3}, \; \frac{1}{3} \right). $$

The center of the circle is at $(2,2,1),$ and the plane goes through the origin, so all we really did was find the two points on the given sphere along that line, $$ x = 2t,y=2t,z=t. $$

Answer 2

If you eliminate, say, $x$ using the second constraint, you are left with a two-variate function and a single constraint: Minimize/Maximize $d^2 = 5z^2 + y^2$ subject to $(y-2)^2 + 5(z-1)^2 = 4$. Now, a Lagrange solution is not bad. From here, you can play also with basic algebra and trig to avoid calculus altogether.

Answer 3

$\newcommand{\e}{\mathbf{e}}$Here's an algebraic approach:

Let $P$ denote the plane with equation $x - 2z = 0$ and $S$ the sphere $$ (x - 1)^{2} + (y - 2)^{2} + (z - 3)^{3} = 9. $$

The vectors $\e_{1} = (2, 0, 1)/\sqrt{5}$ and $\e_{2} = (0, 1, 0)$ are an orthonormal basis of $P$. The orthogonal projection to $P$ of $c = (1, 2, 3)$, the center of $S$, is $$ \langle c, \e_{1}\rangle \e_{1} + \langle c, \e_{2}\rangle \e_{2} = \sqrt{5}\e_{1} + 2\e_{2} = (2, 2, 1) = c_{0}. $$ Since the distance from $c$ to $c_{0}$ is $\sqrt{5}$, the intersection of $S$ and $P$ is a circle of radius $\sqrt{3^{2} - 5} = 2$ centered at $c_{0}$.

In the Cartesian coordinates defined by $\{\e_{1}, \e_{2}\}$, the center $c_{0}$ has coordinates $(\sqrt{5}, 2)$, and a bit of algebra shows the nearest point to the origin is $(\sqrt{5}, 2)/3$, while the furthest point is $5(\sqrt{5}, 2)/3$.

Converting back to spatial coordinates, the nearest point to the origin is $(2/3, 2/3, 1/3)$ and the furthest point is $(10/3, 10/3, 5/3)$.