Can you explain the solution?

Question

Find the extreme values of an implicitly given function z of variables x and y: $x^2+y^2+z^2-xz-yz+2x+2y+2z-2$
Solution:

I don't understand how we got the second equation from the first (given). Then how we got the system from the second?

Answer 1

Consider a geometric representation.

You have a 3D surface with equation

$$f(x,y,z)=0 \tag{1}$$

which is considered (at least locally) as

$$z=\phi(x,y) \tag{2}$$

(function $\phi$ could be explicited by considering (1) as a quadratic in variable $z$ but that's another story)

Now, in each point $(x,y,z)$ of the surface there is a gradient vector (normal vector to the surface) given by

$$\left(\partial f/\partial x, \ \partial f/\partial y, \ \partial f/\partial z \right)$$

like the normal vectors you can see in the answer to this question.

In the system of 3 equations, setting $\partial f/\partial x=0 \ \partial f/\partial y=0$ constraints the gradient to be vertical which is a necessary condition for having an extremum for $z$. The third equation constrains $(x,y,z)$ to belong to the surface.

Answer 2

Implicit differentiation is being used to find where the gradient of the graph is $0$. That will be the extreme value.

Answer 3

In case it is the details that you are asking about, as Jean Marie explained, the solver is writing the expression describing the implicit function $ \ z(x,y) \ $ as $ \ x^2 + y^2 + z^2 - xz - yz + 2x + 2y + 2z - 2 \ = \ 0 \ \ $ to construct a surface in $ \ \mathbb{R}^3 \ \ . $ (This is a "tilted" ellipsoid that is not centered on the origin and has none of its axes parallel to the coordinate axes; see the graph at the end of this post.)

If we differentiate this implicitly with respect to the two independent variables, we obtain $$ \ \frac{d}{dx} \ [ \ x^2 \ + \ y^2 \ + \ z^2 \ - \ xz \ - \ yz \ + \ 2x \ + \ 2y \ + \ 2z \ - \ 2 \ ] \ \ = \ \ \frac{d}{dx} \ [ \ 0 \ ] $$ $$ \Rightarrow \ \ 2x \ + \ 0 \ + \ 2z·\frac{dz}{dx} \ - \ \underbrace{(z \ + \ x·\frac{dz}{dx})}_{\text{by Product Rule}} \ - \ y·\frac{dz}{dx} \ + \ 2 \ + \ 0 \ + \ 2 ·\frac{dz}{dx} \ - \ 0 \ \ = \ \ 0 \ \ $$ $$\Rightarrow \ \ (2z \ - \ x \ - \ y \ + \ 2)·\frac{dz}{dx} \ + \ (2x \ - \ z \ + \ 2) \ = \ \ 0 \ \ \Rightarrow \ \ \frac{dz}{dx} \ \ = \ \ -\frac{2x \ - \ z \ + \ 2}{2z \ - \ x \ - \ y \ + \ 2} \ \ ; $$ $$ \ \frac{d}{dy} \ [ \ x^2 \ + \ y^2 \ + \ z^2 \ - \ xz \ - \ yz \ + \ 2x \ + \ 2y \ + \ 2z \ - \ 2 \ ] \ \ = \ \ \frac{d}{dy} \ [ \ 0 \ ] $$ $$ \Rightarrow \ \ 0 \ + \ 2y \ + \ 2z·\frac{dz}{dy} \ - \ x·\frac{dz}{dy} \ - \ (z \ + \ y·\frac{dz}{dy}) \ + \ 0 \ + \ 2 \ + \ 2 ·\frac{dz}{dy} \ - \ 0 \ \ = \ \ 0 \ \ $$ $$\Rightarrow \ \ (2z \ - \ x \ - \ y \ + \ 2)·\frac{dz}{dy} \ + \ (2y \ - \ z \ + \ 2) \ = \ \ 0 \ \ \Rightarrow \ \ \frac{dz}{dy} \ \ = \ \ -\frac{2y \ - \ z \ + \ 2}{2z \ - \ x \ - \ y \ + \ 2} \ \ . $$ (The solution in the text writes a single equivalent equation using differentials.)

By setting these two derivatives equal to zero, we obtain the two planes in $ \ \mathbb{R}^3 \ \ , \ \ 2x - z + 2 \ = \ 0 \ \ $ and $ \ 2y - z + 2 \ = \ 0 \ \ . $ (The common denominator in the two ratios is not equal to zero at any point on the ellipsoid where the two numerators are.) The intersection of these two planes is the line $ \ x \ = \ y \ = \ \frac12z - 1 \ \ ; $ where this line in turn intersects the ellipsoidal surface are the critical points of the implicit function $ \ z(x , y) \ \ . $ (The last image shows a plot of the ellipsoid and the two planes.)

The solution does not show the calculation, but if we insert $ \ y \ = \ x \ $ and $ \ z \ = \ 2x + 2 \ \ $ into the surface equation, we have the quadratic equation $$ 2x^2 + 12x + 6 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ y \ = \ \ -3 \ \pm \ \sqrt6 \ \ \Rightarrow \ \ z \ \ = \ \ 2x + 2 \ \ = \ -4 \ \pm \ 2·\sqrt6 \ \ . $$ (Alternatively, using $ \ \frac{z}{2} - 1 \ = \ y \ = \ x \ \ $ produces $ \ \frac12z^2 + 4z - 4 \ = \ 0 \ \ $ $ \Rightarrow \ z \ = \ -4 \ \pm \ 2·\sqrt6 \ \ . ) $

So we find the extremal values of the implicit function to be $ \ z_{max} \ = \ -4 + 2·\sqrt6 \ \approx \ +0.899 \ $ and $ \ z_{min} \ = \ -4 - 2·\sqrt6 \ \approx \ -8.899 \ \ . $

In common practice, we would construct a $ \ 3 \times 3 \ $ Hessian matrix of second derivatives to show that the values of $ \ z \ $ we have computed are in fact the minimal and maximal values. (Because of the form of the first derivatives, this would become somewhat complicated.) The solution you have appeared to be demonstrating the convexity of the surfaces at the critical points by using differentials based on second derivatives, which can be written more succinctly.