# How to find local extrema of implicit function $F:\ z^2+xyz-xy^2-x^3=0$ as $z=f(x,y)$?

It is a calculus homework, professor have explained me today how to find extrema if $$F$$ were a sphere, I got it but I cant solve this one. So, I am to tell where I have stopped at:

Find local extrema of $$F:\ z^2+xyz-xy^2-x^3=0$$ as $$z=f(x,y)$$

#### Solving

Let $$z=f(x,y)$$, then $$\frac{\delta z}{\delta x} = - \frac{F_x'}{F_z'}=-\frac{yz-y^2-3x^2}{2z+xy},\\ \frac{\delta z}{\delta y}= - \frac{F_y'}{F_z'}=-\frac{xz-2xy}{2z+xy}.$$ Then, we have to find critical points. To do that we need to solve $$\begin{cases} \frac{\delta z}{\delta x} = 0,\\\ \frac{\delta z}{\delta y}=0.\end{cases}\ \Leftrightarrow \ \begin{cases}\frac{yz-y^2-3x^2}{2z+xy}=0,\\\ -\frac{xz-2xy}{2z+xy}=0. \end{cases}$$ Which is equal to $$yz-y^2-3x^2=xz-2xy.$$

And at this point I am stopped. I don't know how to find the critical points from the system with 2 equations and 3 varriables.

• It's not clear what you mean by $F:\ z^2+xyz-xy^2-x^3=0$ as $z=f(x,y)$. Does it imply $F:\ f(x,y)^2+xyf(x,y)-xy^2-x^3=0$. Or does it mean that F = f(x,y) and z inside f(x,y) is a constant?
– Lu4
Commented May 19, 2023 at 18:54
• @Lu4 I am sorry. I mean that we have some surface $F:\ z^2+xyz-xy^2-x^3=0$ and we should find its extrema along the $z$ axis, like we would do with the $z=x^2+y^2$ function. Commented May 19, 2023 at 19:02
• If $z$ is a function then it's not a variable. Commented May 19, 2023 at 19:56

You can use the Lagrange multipliers as well

Defining

$$L(x,y,z,\lambda) = z + \lambda(z^2 + x y z - x y^2 - x^3)$$

the stationary points are the solutions for

$$\nabla L = 0 = \cases{ \lambda \left(y z-3 x^2-y^2\right)\\ \lambda (x z-2 x y)\\ \lambda (x y+2 z)+1\\ x y z+z^2-x^3-x y^2 }$$

and after solving we have

$$(x,y,z,\lambda) = (-6,\pm 6\sqrt{3},\pm 12\sqrt{3},\pm\frac{1}{12\sqrt{3}})$$

one is a relative minimum and the other a relative maximum.

NOTE

From the Lagrangian conditions, with $$f = f(x,y,z)=0$$

$$\cases{ \lambda f_x=0\\ \lambda f_y= 0\\ 1+\lambda f_z=0\\ f=0 }$$

discarding the case of $$\lambda = 0$$ we have equivalently the conditions

$$\cases{ f_x = 0\\ f_y = 0\\ f=0 }$$

• Hey Cesareo, can you check if those points actually optimize the source function? I got those resulting points not being part of original surface
– Lu4
Commented May 19, 2023 at 21:02
• According to my calculations, the found points pertain to the restriction $z^2 + x y z - x y^2 - x^3 = 0$. Commented May 19, 2023 at 21:12
• The negative solution does look good, there seems to be an issue with the positive one
– Lu4
Commented May 19, 2023 at 21:19
• I guess the points are $(-6,6\sqrt{3},12\sqrt{3}),(-6,-6\sqrt{3},-12\sqrt{3})$
– Lu4
Commented May 19, 2023 at 21:24
• OK. You are right. I forgot to consider the $x$ signal which remains negative Commented May 19, 2023 at 21:24

I've plotted your function to better understand its behavior. It may be beneficial for you to reassess your problem's definition in light of these results.

What we observe is that your function describes two surfaces along the z-axis that extend towards both positive and negative infinity. This implies that if you're seeking extrema along this axis, you'll only find one, either a minimum or a maximum depending on your direction of search.

However, the existence of these two infinite surfaces complicates the matter of finding a globally optimal solution. The 'optimality' of the extremum you find will always be relative to the surface you're currently exploring, and it will not necessarily correspond to the overall minimum or maximum of the function.

This is because the second surface, not being evaluated in your search, will always extend further towards infinity. Consequently, you'll always have an unexplored area of the function that could potentially yield a more extreme value.

In summary, the peculiar behavior of your function, with its dual infinite surfaces, means that a truly optimal solution — in the sense of finding an absolute maximum or minimum — may not be possible. Instead, your findings will always be dependent on the specific surface you're examining.