# Is it possible to explicitly determine max and min in this case?

On the set

$$A = \{(x, y, z) \in \mathbb{R}^3; xy + xz + yz \leq 1\}$$

does the following function admit global/local max/min?

$$f(x, y, z) = xyz$$

attempts

So, firt of all I proved we cannot appeal to Weierstrass Theorem, for the set $$A$$ is closed but not bounded.

It's closed because set of the form $$[a, +\infty)$$ (in one dimension) are closed. It's not bounded because, say, restricting on $$x = y$$ and $$x = z$$, I obtain the restricted condition $$- z^2 \leq 1$$, hence $$z$$ can be large enough such that no ball $$B_r(0)$$ can contain $$A$$.

(I proved it more rigorously; I omit this part because it's not important for the question).

Now, for some reason I understood that $$xy + xz + yz \leq 1$$ represents a so called elliptic hyperboloid (when we take the $$=$$ sign, thanks to GeoGebra). I really do not know how to reduce that equation into the canonical form for an elliptic hyperboloid which shall be

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = - 1$$

(some help about would be appreciated, even just to understand the method, but again: not really fundamental now).

Since Weierstrass does not hold, there is no certainty about the existence of global max/min over $$A$$.

• Question: is there a way to find those max/min, or to completely exclude their existence (or to prove it)?

I thought I could study the behaviour of $$f$$ in the internal points of $$A$$ (with the gradient of $$f$$) and the points on the boundary.

$$\nabla f = (0, 0, 0) \rightarrow \begin{cases} yz = 0 \\ xz = 0 \\ xy = 0 \end{cases}$$

Which are satisfied for points of the form $$(0, 0, z)$$ or $$(0, y, 0)$$ or $$(x, 0, 0)$$.

In all those points, the function returns zero as a value.

I am stuck however on the boundary $$\partial A$$, because I am not able to write down the equation for $$\partial A$$. If I could, I would study the restriction of $$f(x, y, z)$$ on the boundary, and maybe manage a bit the thing.

So is there a more efficient way? How to write down $$\partial A$$?

• The boundary is just the set with the $=$ instead of the $\le$ right? Commented Mar 2, 2023 at 22:10
• @student91 Uh, well actually in this case yes! But how then to restrict the function to the boundary? Shall I solve for one of the variables and substitute? Commented Mar 2, 2023 at 22:10

Well it has been a while, so I am not 100% sure. I think there isn't a maximum.

proof:

let $$\epsilon > 0$$ and choose $$x = -\epsilon, z = -y , y >1$$ then $$f(x,y,z) = xyz = (-\epsilon)(-y)(y) = \epsilon y^2 > \epsilon$$ on the other hand $$xy + xz + yz = xy + x(-y) + y(-y) = -y^2 \leq 0 \leq 1$$.

I guess minimum goes analogue.

• How about local min/max? Commented Mar 2, 2023 at 22:22
• But most of all: why the proposed method is wrong? I mean, I resolve for z $$z = \frac{1-xy}{x+y}$$ and restrict $f$ to that, getting a two variable function $$f(x, y) = xy \frac{1-xy}{x+y}$$ Then I study the gradient, obtaining two solutions $$\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$$ and the identical positive one. I mean it's like in a two variables function $f(x, y)$ when I study internal points through the gradient, then for the boundary I restrict $f$ on the boundary and then evaluating $f'$ I find the points on the boundary. Eventually I compare them all. Commented Mar 2, 2023 at 22:34
• Perhaps since the constraint is not bounded, what I will find will be local min/max? Commented Mar 2, 2023 at 22:34
• I am not sure what you are trying to proof. Proofing that a theorem is not applicable won't ever help you, that just states nothing. Commented Mar 2, 2023 at 22:42
• I don't think you understood what I am doing here, but I'm pretty sure I'm right. I also checked with W. Mathematica, and the two points I get are indeed local max / local min. Commented Mar 2, 2023 at 22:43

I answer my own question in a rapid way, but I got it.

• Find the boundary as $$xy + yz + zx = 1$$

• As computed above, the gradient is correct, and those points are eventual candidates for loval max/min

• on the boundary, restrict $$f$$. Solve for example for $$z$$ and get a two variables function $$f(x, y)$$

• To study the points on the boundary, compute the gradient of that function, solve it and you get two distinct points

• Find the value for $$z$$ in those cases

• Evaluate $$f(x, y, z)$$ on all the points, and conclude.

Here is a (perhaps) more systematic way to handle such problems.

The critical points of a multivariate function over a closed domain can lie on its interior or its boundary.

The first ones are calculated "as usual" and as you did, i.e. by setting the gradient to zero. Here we have $$f(x,y,z) = xyz$$, hence $$\nabla f(x,y,z) = (yz,zx,xy) = (0,0,0)$$, which leads to $$(x_0,y_0,z_0) = (0,0,0)$$.

On the boundary $$\partial A = \{(x,y,z)\in\mathbb{R}^3 \,|\, xy+yz+zx=1\}$$, we will face an optimization under constraint, that is why we need to use the method of Lagrange multipliers. In pratice, the new function to optimize is given by the Lagrangian $$L(x,y,z,\lambda) = f(x,y,z) - \lambda(xy+yz+zx-1)$$. One has the following system of equations : $$\begin{cases} \partial_x L = yz-\lambda(y+z) = 0 \\ \partial_y L = zx-\lambda(z+x) = 0 \\ \partial_z L = xy-\lambda(x+y) = 0 \\ \partial_\lambda L = xy+yz+zx-1 = 0 \end{cases}$$ whose solutions are given by $$(x_\pm,y_\pm,z_\pm,\lambda_\pm) = \pm\frac{1}{\sqrt{3}}(1,1,1,1/2)$$.

The three stationary points we found produce the following values : $$f(x_0,y_0,z_0) = 0$$ and $$f(x_\pm,y_\pm,z_\pm) = \pm3^{-3/2}$$, such that $$(x_0,y_0,z_0)$$ is a saddle-point, while $$(x_+,y_+,z_+)$$ turns out to be a maximum and $$(x_-,y_-,z_-)$$ a minimum.

N.B. : instead of computing the images of the critical points, it is also possible to study the eigenvalues of the Hessian matrix of $$f$$ at those points in order to distinguish the maxima/minima/saddle-points. In the present case, this matrix is given by $$\mathrm{Hess}(f)(x,y,z) = \begin{pmatrix} 0 & z & y \\ z & 0 & x \\ y & x & 0 \end{pmatrix}$$

Addendum. It is also possible to apply a version of Lagrange multiplier method which is adapted to inequality constraints, so that you can find all the critical points on $$A$$ at the same time.

The condition $$xy+yz+zx\le1$$ means that the quantity $$xy+yz+zx-1$$ is a negative number, that we will call $$-t^2$$, such that the inequality constraint now becomes an equality constraint of the form $$xy+yz+zx-1 = -t^2 \le 0$$.

The Lagrangian is then $$\mathscr{L}(x,y,z,\lambda) = f(x,y,z) - \lambda(xy+yz+zx-1+t^2)$$. The same optimization procedure as above leads to the solution $$x_* = y_* = z_* = 2\lambda_* = \pm\sqrt{\frac{1-t^2}{3}}$$, which is only defined for $$|t|\le1$$, hence $$f(x_*,y_*,z_*) = \pm\left(\frac{1-t^2}{3}\right)^{3/2}$$. The case $$t=0$$ corresponds to the critical points lying on $$\partial A$$ $$-$$ we recognize the two solutions derived above $$-$$, whereas the cases where $$t\neq0$$ are associated to the points belonging to the interior $$\mathring{A}$$ and can be studied as if $$f$$ was a standard function of $$t$$.