Dtyjlui

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

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

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

I found out $(0,pmfrac{5}{2},pmfrac{sqrt{39}}{2}),(pmfrac{sqrt{39}}{2},pmfrac{5}{2},0),(0,0,pm4)$

? 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

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

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

$ 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 = 2xlambda e^y $

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

$(3) 2z = 2zlambda 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,-3sqrt{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 $ yin(-sqrt{3},sqrt{3})$, and $g(y) > 0 $ if $ yin(-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 ).