Extremas using Lagrange or other methods with exponential functions $ \overline{B((0,0),1))}\mapsto \mathbb{R}$
$f(x,y)= e^{(x-1)y}$
How do you find the extremas for this functions? I tried using Lagrange method but I have trouble solving the system of equations with exponential.
My border is the circle $x^2+y^2=1$ So I have to check on interior points doing the gradient I find a stationary point at (1,0) but that is on the border so I have to check for extremas on the border. But I am stuck with lagrange. $\nabla f = \lambda \nabla g$ where $g=0$ is the circle.
Is there a simpler way to do it?
 A: Your method is correct. To find the extremas of $f$ on the boundary $g=0$ we can proceed as follows
Method 1: Lagrange multipliers.
The necessary condition as you mentioned is $\nabla f=\lambda\nabla g$ (or $\nabla g=0$, but this is not the case for $g=0$), that is
$$
\begin{cases}
e^{(x-1)y}y=2\lambda x,\\
e^{(x-1)y}(x-1)=2\lambda y
\end{cases}\quad\Rightarrow\quad x\ne 0\text{ and } \frac{2\lambda}{e^{(x-1)y}}=\frac{y}{x}=\frac{x-1}{y}\quad\Rightarrow\quad y^2=x^2-x.
$$
Using $x^2+y^2=1$ we get 
$$
x^2+x^2-x=1\quad\Leftrightarrow\quad 2x^2-x-1=(x-1)(2x+1)=0
$$
which gives three candidates for the extremas $(1,0)$, $(-1/2,\pm\sqrt{3}/2)$. Compare the values of $f(x,y)$ there and conclude.
Remark 1: We can do a bit smarter here. Hence the exponential is an increasing function we can look for extremas for $F(x,y)=(x-1)y$ instead.  The points of maximum and minimum will be the same.
Remark 1: In case of two variables $(x,y)$ the necessary condition $\nabla f\|\nabla g$ is equivalent to $\det[\nabla f\ \nabla g]=0$ which does not have $\lambda$ variable
$$
\left|\begin{matrix}e^{(x-1)y}y & 2x\\e^{(x-1)y}(x-1) & 2y\end{matrix}\right|=2e^{(x-1)y}(y^2-(x-1)x)=0\quad\Leftrightarrow\quad y^2=x^2-x.
$$
Method 2: Direct parameterization of the boundary.
Since $x^2+y^2=1$ can be parameterized as
$$
\begin{cases}
x=\cos t,\\
y=\sin t
\end{cases}
$$
for $t\in[0,2\pi]$ the problem is to minimize/maximize 
$$
h(t)=f(\cos t,\sin t)=e^{(\cos t-1)\sin t},\quad t\in[0,2\pi]
$$
which can be done by simple differentiation and solving $h'(t)=0$. Again we can replace the function $h$ with $\tilde h(t)=(\cos t-1)\sin t$.
