find extrema of $f(x,y,z)=z$ with domain

Question

$D=\{(y^2+z^2)/6\le x,x^2+y^2=z^2+16\}$ in $\mathbb R$

on $f:D \to R ,f(x,y,z)=z$

• First off just with a brief look at my function I can say that there are no critical points ($f_x=0,f_y=0,f_z=1$), (morover its an open set which means that the max and min cannot occur ? correct me if I'm wrong)
• I can now analyze the domain $D$, what I can say is that :

$(y^2+z^2)/6\le x\Longrightarrow$ It's a kind of Paraboloid

$x^2+y^2=z^2+16 \Longrightarrow$ It's a Hyperboloid of One Sheet

If I want to find the boundary of $D$, I need to put those two equations in a system, finding: $x^2-2z^2+6x-16=0$ which is the intersection between those two surfaces.

• Now in order to find possible min/max points, I can use the Lagrange multiplier system with $x^2-2z^2+6x-16=0$ as a constraint.

$$\lambda (2x+6)=0$$

$$0=0$$

$$1+\lambda (-2z)=0$$

$$x^2-2z^2+6x-16=0$$

• From the first equation I can say that it is TRUE for $\lambda = 0$ or $x=-3$ , but $\lambda$ cannot be zero becouse It doesn't satisfy the third equation . What I can do instead is using $x=-3$ in the forth equation , but here is the problem : I remain with $-z^2=25$ and I conclude that I didn't find any points. Even if I use the rhird equation finding $z$ and putting it inside the forth equation it still gives me something like $\lambda^2 $ equal to a negative number.

• Where did I make the mistake? (maybe the boundary ?)

Answer 1

The domain $D$ is a two-dimensional wobbly disc cut out from the hyperboloid. The boundary of this disc is the curve $\gamma$ of intersection between the hyperboloid and the paraboloid. This curve is determined by two constraints (and not by one, as you seem to believe).

Drawing a figure showing the projection to the $(x,z)$-plane suggests that the point on $\gamma$ with maximal $z$-coordinate is on the symmetry plane $y=0$. There are two such points, namely $P_\pm:=\bigl(8,0, \pm4\sqrt{3}\bigr)$. These two points would then lead to the extremal values $\pm4\sqrt{3}$ of $f\restriction D$.

Doing it the hard way we first remark that the hyperboloid has no horizontal tangent planes. It follows that necessarily the extrema of $f\restriction D$ are taken on $\partial D=\gamma$. We therefore set up the Lagrangian $$\Phi:=z-\lambda(6x-y^2-z^2)-\mu(x^2+y^2-z^2-16)$$ and then obtain the equations $$\eqalign{\Phi_x&=-6\lambda-2\mu x=0 \cr \Phi_y&=2\lambda y-2\mu y=0\cr \Phi_z&=1+2\lambda z+2\mu z=0\ .\cr}$$ From the second equation it follows that $y=0$ or $\lambda=\mu$. The case $y=0$ leads to the points $P_\pm$ we have already found. When $\lambda=\mu$ we are left with the equations $$ \lambda(3+x)=0,\quad 4\lambda z=-1\ .$$ This enforces $\lambda\ne0$, hence $x=-3$. But there are no points in $D$ with $x=-3$.

To sum it up: The extrema of $f\restriction D$ are taken in the points $P_\pm$.

Answer 2

Using a slack variable $\epsilon$ to convert the inequality into an equality we have

$$ L(z,y,z,\lambda,\mu,\epsilon) = z+\lambda\left(\frac 16(y^2+z^2)-x+\epsilon^2\right)+\mu(x^2+y^2-z^2-16) $$

the stationary points are the solutions for

$$ \nabla L = \left\{ \begin{array}{rcl} 2 \mu x-\lambda &=&0 \\ \frac{\lambda y}{3}+2 \mu y&=&0 \\ \frac{\lambda z}{3}-2 \mu z+1&=&0 \\ \epsilon ^2-x+\frac{1}{6} \left(y^2+z^2\right)&=&0 \\ x^2+y^2-z^2-16&=&0 \\ 2 \epsilon \lambda &=&0 \\ \end{array} \right. $$

which are

$$ \left[ \begin{array}{cccccc} x & y & z & \lambda & \mu & \epsilon \\ 8 & 0 & -4 \sqrt{3} & \frac{2 \sqrt{3}}{5} & \frac{\sqrt{3}}{40} & 0 \\ 8 & 0 & 4 \sqrt{3} & -\frac{2 \sqrt{3}}{5} & -\frac{\sqrt{3}}{40} & 0 \\ \end{array} \right] $$

As we can observe the two stationary points are with $\epsilon = 0$ meaning that the restriction is actuating and they are at the feasible region border.

Attached a plot showing in red the extrema locations as well as the intersection curve between the borders of the two restrictions in blue.

Attached the MATHEMATICA script to produce the plot h = (y^2 + z^2)/6 - x; g = (x^2 + y^2 - z^2 - 16); p1 = {8, 0, -4 Sqrt[3]}; p2 = {8, 0, 4 Sqrt[3]}; gr1 = Graphics3D[{Red, Sphere[p1, 0.3]}]; gr2 = Graphics3D[{Red, Sphere[p2, 0.3]}]; gr0 = ContourPlot3D[{h == 0, g == 0}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, MeshFunctions -> {Function[{x, y, z, f}, h - g]}, MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}}, ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]], PlotPoints -> 40]; Show[gr0, gr1, gr2, PlotRange -> All] `