# Minima maxima for a 3 variable function on a whole Critical line ( and not a point)

Say if i have a function

$f(x,y,z)= xyz(16-x-y-2z)$

and i am looking for maxima and minima for it.

With a quick calculation and after demanding that the

$\nabla f = 0$

We get that the critical values are $(4, 2 , 2)$ and the maix axes lines aswell: $(x,0 , 0 ) ; (0,y,0) ; (0,0,z)$

for the point $(4,2,2)$ i can use the Hessian matrix and check ( and i think it turns out as a Maxima)

But im not sure what to do with the lines... The hessian matrix does not work for it ( caus it is equal 0 at thoes lines)

and i am not sure how to check and detrmin for them theire type.

Thank you very much!

Although it does not affect the discussion materially, I believe your maximum is located at $\ ( 4, \ \mathbf{4}, \ 2 ) \$ , as the first partial derivatives of the function are

$$f_x \ = \ yz \ (16 \ - \ 2x \ - \ y \ - \ 2z) \ \ ,$$ $$f_y \ = \ xz \ (16 \ - \ x \ - \ 2y \ - \ 2z) \ \ , \ \ \text{and}$$ $$f_z \ = \ xy \ (16 \ - \ x \ - \ y \ - \ 4z) \ \ .$$

Because the exponents of any of the variables in $\ f( x, \ y, \ z) \$ are only as high as two, the second partial derivatives are not all zero, so the Hessian does have non-zero entries. Unfortunately, that doesn't help matters because the eigenvalues are zero for any of the matrices evaluated along each of the coordinate axes. For example, the matrix for $\ ( x, \ 0, \ 0) \$ becomes

$$\left[ \begin{array}{ccc}\ 0&0&0\\0&0&16 - x^2\\0&16 - x^2&0\end{array} \right] \ \ .$$

(Without working everything out here, we find the Hessian at $\ ( 4, \ 4, \ 2 ) \$ to be, by contrast, $\left[ \begin{array}{ccc}\ -16&-8&-16\\-8&-16&-16\\-16&-16&-64\end{array} \right]$ , which does have three negative eigenvalues, confirming that it is a relative maximum.)

What is left for us in situations such as these is to take "slices" through the "surface" of the function to see what its behavior is in the neighborhood of points of interest. That becomes a bit difficult to "visualize" when we have a function of three variables, so it sometimes takes a bit of experimentation to find satisfactory "views" of the function in order to learn about its character.

Here, we can explore what happens along the $\ x-$ axis by using the "slices" $\ z \ = \ y \$ and $\ z \ = \ -y \$ . For specific choices $\ x \ = \ X \$ , our function becomes

$$f( X, \ y, \ y) \ = \ X \ y^2 \ ( [16 - X] \ - \ 3y) \ = \ X \ (16 - X) \ y^2 \ - \ 3Xy^3$$ and $$f( X, \ y, \ -y) \ = \ -X \ y^2 \ ( [16 - X] \ + \ y) \ = \ -X \ (16 - X) \ y^2 \ + \ Xy^3 \ \ .$$

Close to points $\ (X, \ 0, \ 0) \$ on the $\ x-$ axis, the quadratic term is dominant, so we have

$$f( X, \ y, \ y) \ \sim \ X \ (16 - X) \ y^2 \ \ \ \text{and} \ \ \ f( X, \ y, \ -y) \ \sim \ -X \ (16 - X) \ y^2 \ \ .$$

On the $\ z \ = \ y \$ slice, the surface for this function is then "concave downward" for $\ X \ < \ 0 \$ and $\ X \ > \ 16 \$ , and "concave upward" on $\ 0 < X < 16 \$ . However, on the $\ z \ = \ -y \$ slice, we see that all of these concavities are reversed on the same intervals. Therefore, all of the points $\ (X, \ 0, \ 0) \$ will be saddle points, with the possible exceptions of $\ (0, \ 0, \ 0) \$ and $\ (16, \ 0, \ 0) \$ . (The investigation would become a bit more complicated at those locations, though I suspect they too are saddle points.)

A similar analysis conducted for points along the other two coordinate axes produces analogous conclusions.

[It may be of interest to look at a couple of related problems, one with a somewhat more complicated function, which present much the same situation, to see more of this sort of investigation.]