Question

The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest to and farthest from the origin. nearest point (x, y, z) = farthest point (x, y, z) =

Answer 1

The distance between a point $(x,y,z)$ and the origin is $\sqrt{x^2+y^2+z^2}$. But since $(\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}$, both $f(x)$ and $\sqrt{f(x)}$ have the same critical points, so we can consider instead the squared distance, $x^2+y^2+z^2$.

We're looking for the extrema of $x^2+y^2+z^2$ subject to $x+y+2z=12$ and $z=x^2+y^2$. The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)$

with critical points where the partial derivatives vanish:

$L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)$

$L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)$

$L_z=2z+2\lambda+\mu=0$

$L_\lambda=x+y+2z-12=0$

$L_\mu=z-x^2-y^2=0$

From the first two equations, it follows that $x=y$.

Then in the last two equations,

$x+y+2z-12=0\implies x+z=6$

$z-x^2-y^2=0\implies z=2x^2$

$\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2$

If $x=\frac32$, then $z=6-\frac32=\frac92$; if $x=-2$, then $z=8$.

So there are two critical points, $\left(\frac32,\frac32,\frac92\right)$ and $(-2,-2,8)$.

Let $f(x,y,z)=\sqrt{x^2+y^2+z^2}$. We have a minimum distance of $f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2}$ and maximum distance of $f(-2,-2,8)=\boxed{6\sqrt2}$.