How to find extrema for given implicit function of $(x,y,z)$ variables?

Question

Given the equation $$z^3 - 33xyz - 27 = 0,$$ we know that $z=z(x,y)$. As $z$ is function of both $x$ and $y$, I don't really know how to search for extremeum, because of three variables. I only solved for two variables. I tried with wolfram but some kind of skate park was a graphic solution.

Thanks for any help.

Answer 1

Do you know the implicit function theorem ?

Setting

$$g:(u,v,w) \to w^3 -33uvw -27 $$

Requirement

Visit Introduction to implicit function theorem

Answer 2

Without some sort of restriction on the domain of $ \ z \ $ , the function has no global maximum or minimum. In the equation $ \ z·(z^2-33xy) = 27 \ \ , $ allowing the factor $ \ z^2 - 33xy \ $ to approach zero then permits $ \ z \ $ to increase or decrease without limit. The surface representing the implicit function then asymptotically approaches the quasi-hyperbolic cylinder $ \ z^2 = 33xy \ $ which extends infinitely in either vertical direction. So the function has no global extrema.

The function also asymptotically approaches the $ \ xy-$plane, since for $ \ xy > 0 \ $ , allowing the product to grow without limit causes $ \ z \ $ to approach zero "from below" ( $ z^2 - 33xy \ $ being negative), while for $ \ xy < 0 \ $ , the function approaches zero "from above".

Using implicit differentiation, we can produce partial derivatives $$ \frac{\partial }{\partial x} \ [ \ z^3 \ - \ 33xyz \ ] \ = \ \frac{\partial }{\partial x} \ [ \ 27 \ ] \ \ \Rightarrow \ \ 3z^2 · \frac{\partial z }{\partial x} \ - \ 33yz \ - \ 33xy · \frac{\partial z }{\partial x} \ = \ 0 $$ $$ \Rightarrow \ \ \frac{\partial z }{\partial x} \ = \ \frac{11yz}{z^2 \ - \ 11xy} \ \ , \ \ \text{and similarly} \ \ \frac{\partial z }{\partial y} \ = \ \frac{11xz}{z^2 \ - \ 11xy} \ \ . $$ Since $ \ z \ $ never attains the value of zero, the first partial derivatives both equal zero only at the single critical point at $ \ x = y = 0 \ \ , $ where $ \ z^3 = 27 \ \Rightarrow \ z = 3 \ \ . $ In the neighborhood of $ \ (0,0,3) \ \ , $ the surface in the plane $ \ y = x \ $ is "concave upward" [ $ \ z·(z^2-33x^2) = 27 \ $ ] , while it is "concave downward" in the plane $ \ y = -x \ $ [ $ \ z·(z^2+33x^2) = 27 \ $ ] . Thus, $ \ (0,0,3) \ $ is a saddle point. [Hence, your "skate park"...]

Views of the surface [in light blue] for $ \ z = f(x,y) \ $ appear below, with the asymptotic surface $ \ z^2 = 33xy \ $ shown in yellow.