find extrema of a $f(x,y,z)$ function using Lagrange multiplier

Question

The function is : $f(x,y,z)=e^y(x^2+z^2)$

restricted on $R=\{x^2-3y^2+z^2+9=0,x^2+y^2+z^2\le 16\}$

$$ \left\{ \begin{aligned} 2xe^y=\lambda 2x+\mu 2x \\ e^y(x^2+z^2)=-\lambda 6y+\mu 2y\\ 2ze^y=\lambda 2z+\mu 2z\\ x^2-3y^2+z^2+9=0\\ x^2+y^2+z^2-16=0\\ \end{aligned} \right. $$

I found out $(0,\pm\frac{5}{2},\pm\frac{\sqrt{39}}{2}),(\pm\frac{\sqrt{39}}{2},\pm\frac{5}{2},0),(0,0,\pm4)$

? are they right ?I don't think so becouse using the second derivate test It seems that they are all minimums points.

and the image of the function $f(R)=[\frac{39}{4}e^{-\frac{5}{2}},16e^{\frac{5}{2}}]$

I also did study the function itself:

$$ \left\{ \begin{array} 2xe^y=0 \\ e^y(x^2+z^2)=0\\ 2ze^y=0 \end{array} \right. $$

so It's seems like there is also the point $(0,y,0)$ ? I don't get it

Answer 1

$x^2 + z^2 = 3y^2 - 9\\ 4y^2 - 9 \le 16\\ y^2 \le \frac {25}{4}\\ |y| \le \frac 52\\ 0\le x^2 + z^2 \le \frac {75}{4} - 9\\ 0 \le x^2 + z^2 \le \frac {39}{4}$

$ 0 \le f(x,y,z) \le e^{\frac 52}(\frac {39}{4})$

Answer 2

$ f:R^3 \to R $

$ f(x,y,z) = e^y(x^2+z^2) $

$ M = \{(x,y,z) \in R^3 : x^2 - 3y^2 + z^2 +9 = 0,\ \ x^2 + y^2 + z^2 \leq 16 \} $

Let's define $ F: R^3 \to R $, $\ \ \ $ $ F(x,y,z) = x^2 - 3y^2 + z^2 + 9 $

$ \nabla F(x,y,z) = [2x,-6y,2z] $, $ \nabla f(x,y,z)=[2xe^y,e^y(x^2+z^2),2ze^y] $

Points where $\nabla F$ is zero vector aren't in M, so everything good so far. We get 4 equations:

$(1) \ \ \ \ \ 2x = 2x\lambda e^y $

$(2) \ \ \ \ \ -6y = \lambda e^y(x^2+z^2) $

$(3) \ \ \ \ \ 2z = 2z\lambda e^y $

$(4) \ \ \ \ \ x^2 - 3y^2 + z^2 +9 =0 $

By simple operations: ( I'll get rid of (4), putting $x^2+z^2 = 3y^2 - 9 $ in (2))

$(1') \ \ \ \ \ x(1-\lambda e^y) = 0 $

$(2') \ \ \ \ \ -6y = \lambda e^y(3y^2 - 9) $

$(3') \ \ \ \ \ z(1-\lambda e^y) = 0 $

Okay, right now, either $ \lambda e^y = 1 $, but then $ 3(y^2 + 2y - 3) = 0 $ which is impossible since $ y \in [-4,4] $ due to inequality in M, $ y \leq 0 $ due to (2) (when we put $ \lambda e^y = 1 $ we get $ -6y = x^2 + z^2 $, so $y \leq 0 $, and since (4) we get $y \leq -\sqrt{3} $, so putting those together:

if $\lambda e^y = 1$, then it must holds that $ y \in [-4,-3\sqrt{3}]$, but then $y^2 +2y - 3 = 0$, and only root is $y =-3$, but then by (2) , we get $12 = x^2 + z^2$ and so $x^2 + y^2 + z^2 = 18 \geq 16$, so we're not interested in that.

So by that, it does must hold, that $x=0$ and $z=0$ ( cause $\lambda e^1 \neq 1 $, so to have both (1) and (3) we must have $x=z=0$)

Okay, now a little bit easier knowing that, we get:

$3y^2 - 9 = 0 $ which holds iff $ y \in \{-\sqrt{3},+\sqrt{3}\}$

So we get only points of the form $(0,\pm \sqrt{3},0)$

Looking at the function $ g: R - > R \ \ \ g(y) = 3e^y(y^2-3) $ ( that is your f, but restricted to $x^2+z^2 = 3y^2-9$ ), we see, those are neither minima nor maxima points ( $ g(y) < 0 \ $ if $ \ y\in(-\sqrt{3},\sqrt{3})$, and $g(y) > 0 \ $ if $ \ y\in(-\infty,-\sqrt{3}) \cup (\sqrt{3},+\infty) $ ( saddle points ).

The rest of the exercise goes as you did ( so checking the boundary of M when inequality becomes in fact equality ).

Answer 3

In a geometrical interpretation, the surface $ \ x^2 - 3y^2 + z^2 + 9 \ = \ 0 \ \rightarrow -x^2 + 3y^2 - z^2 \ = \ 9 \ $ is a hyperboloid of two sheets with the $ \ y-$ axis as its symmetry axis. As the second constraint is the ball $ \ x^2 + y^2 + z^2 \ \le \ 16 \ \ , $ we are interested in the disjoint portions of the hyperboloid found in the interior of the ball and the circular intersections at its surface.

[The image at left above shows the geometry, sighting toward the origin in the direction of the negative $ \ x-$ axis; the image at right is a view toward the origin from the first octant to allow the intersection circles to be seen. The "plotter limits" are set so that the sphere is "cut open" to permit its interior to be seen.]

The function $ \ f(x,y,z) \ = \ e^y · (x^2 + z^2) \ $ is also symmetrical about the $ \ y-$ axis. This suggests that points at which the extrema occur may have the form $ \ (\pm x\ , \ y \ , \pm x) \ \ . $ But in fact, the axial symmetry of the constraint surface means that all points on a circle $ \ y = K \ , \ x^2 + z^2 = 3·(K^2 - 3) \ $ are on the same level surface for the function.

Where this is an issue for the Lagrange method is that the system of equations $$ x · (e^y \ - \ \lambda \ - \ \mu) \ = \ 0 \ \ , \ \ e^y· (x^2 + z^2) \ = \ 2y·(\mu \ - \ 3\lambda) \ \ , \ \ z · (e^y \ - \ \lambda \ - \ \mu) \ = \ 0 \ \ $$

presents the appearance that single values of $ \ x = 0 \ $ and $ \ z = 0 \ $ will describe the solutions adequately; the method can be misleading when such symmetries are involved.

The three first partial derivatives equated to zero indicate that $ \ x = 0 \ , \ z = 0 \ , $ and $ \ x^2 + z^2 \ = \ 0 \ $ (since $ \ e^y \ \neq \ 0) \ $ which means the entire $ \ y-$ axis is critical for this function; we see that this is the case since $ \ f(x,y,z) \ \ge \ 0 \ , $ so $ \ f(0,y,0) \ = \ 0 \ \ , $ which is the absolute minimum for the function. As to where this occurs on the constraint surface, the "vertices" of the hyperboloid lie at $ \ 3y^2 \ = \ 9 \ \Rightarrow \ (0 \ , \ \pm \sqrt3 \ , \ 0 ) \ . $

Employing the axial symmetry of the surface and the function, thereby taking "vertical cross-sections", we can investigate $ \ f(y) \ = \ e^y · (3y^2 - 9) \ \ $ . As a function of a single variable, we obtain local extrema from $$ \ f'(y) \ = \ e^y · (3y^2 + 6y - 9) \ = \ 3e^y · (y+3) · (y-1) \ = 0 \ \ . $$

However, since the hyperboloid is $ \ x^2 + z^2 \ = \ 3·(y^2-3) \ \ , $ $ \ y = -3 \ $ gives $ \ x^2 + z^2 \ = \ 18 \ \ , $ which is completely outside the sphere, and we have already established that $ \ y = 1 \ $ is between the vertices of the hyperboloid. We are forced then to satisfy the constraints by taking values of $ \ y \ $ that will give us not local, but rather "boundary" extrema.

The exponential factor (and the absence of any other critical points) indicates that an extreme value will occur for the smallest and largest values of $ \ y \ $ permissible, which are found at the intersection circles given by $$ x^2 - 3y^2 + z^2 + 9 \ = \ x^2 + y^2 + z^2 - 16 \ \ \Rightarrow \ \ 4y^2 \ = \ 25 \ \ \Rightarrow \ \ y \ = \ \pm \frac52 \ \ . $$

The intersection circles are then $ \ y = -\frac52 \ , \ x^2 + z^2 = 3·\left( \left[-\frac52 \right]^2 \ - \ 3 \right) \ = \ \frac{39}{4} \ \ , $ on which our function has the value $ \ e^{-5/2}·\frac{39}{4} \ \ , $ and $ \ y = +\frac52 \ , \ x^2 + z^2 \ = \ \frac{39}{4} \ \ , $ where the function has the absolute maximal value $ \ e^{5/2}·\frac{39}{4} \ \ $ everywhere on it.

Answer 4

Another approach

Given the lagrangian

$$ L(x,y,z,\lambda,\mu) = (x^2+z^2)e^y +\lambda(x^2 - 3 y^2 + z^2 + 9)+\mu(x^2 + y^2 + z^2 - 16) $$

as $\lambda,\mu$ are generic multipliers, the lagrangian

$$ L'(x,y,z,\lambda',\mu') = (x^2+z^2) +\lambda'(x^2 - 3 y^2 + z^2 + 9)+\mu'(x^2 + y^2 + z^2 - 16) $$

with $\lambda'=\lambda e^{-y}$ and $\mu'=\mu e^{-y}$, has the same stationary points.

Now calling $X = x^2, Y = y^2, Z = z^2$ this problem can be stated as

$$ \{\min,\max\}\ \ X + Z\ \ \text{s. t.}\ \ \cases{X-3Y+Z-9=0\\ X+Y+Z\le 16\\ X\ge 0\\ Y \ge 0\\ Z\ge 0} $$

which is a linear programming formulation.