Solve this Lagrange system with $f(x,y,z)=e^x$

Question

I have $f(x,y,z)=e^x$

and its constriction as follow: $g(x,y,z)=x^2+z^2-2=0$

$f_x = e^x$, $g_x = 2x$

$f_y = 0$, $g_y = 0$

$f_z = 0$, $g_z = 2z$

The system is:

$e^x +\lambda 2x = 0$

$0 = 0$

$\lambda 2z = 0$

$x^2+z^2-2 = 0$

My thought process is $\rightarrow$ I see that the third equation is TRUE for $\lambda=0$ or $z=0$, I then put $z=0$ in the fourth equation giving: $x^2 = 2$ so $x=\pm\sqrt{2}$

from here I tend to say that I found a point in $(x,y,z)=(\pm\sqrt{2},0,0)$, I'd infer that ($(\sqrt{2},0,0)$) is a maximum and ($(-\sqrt{2},0,0)$) is a minimum. I know that I did something wrong (moreover I didnt even touched the first equation)

• So can you help me understand how to solve this system and giving me your thought process(not just a straight resolution!)?
• in this case the second equation is $0=0$ , what does it mean ? does It give me some useful informations?

Answer 1

From the third equation, what you get is that $\lambda=0$ or that $z=0$. But $\lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=\pm\sqrt2$. So, the solutions of the system are the points of the form $\left(\pm\sqrt2,y,0\right)$, with $y\in\mathbb R$. Among these points, $f$ attains its maximum ($e^{\sqrt2}$) at the points of the form $\left(\sqrt2,y,0\right)$ and the minimum $(e^{-\sqrt2}$) at the points of the form $\left(-\sqrt2,y,0\right)$.

Answer 2

You also had to check whether $\lambda = 0$ is a valid solution. It is not because plugging in $\lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.

This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.

Therefore, $x= \pm\sqrt2$ and $z = 0$ so the extremal points are $(\pm\sqrt{2}, y, 0)$ for $y \in \mathbb{R}$.