Lagrange Multiplicator: find minima and maxima I've the following function: $$ f(x,y) = (x+1)^2 e^y $$ with the additional condition $$ 2(x-1)^2 + y^2 = 3 $$ What I've done so far: I constructed one single function: $$ L(x, y, \lambda) = (x+1)^2 e^y + \lambda(2(x-1)^2+y^2-3)$$ Then, I calculated the derivations: $$ \frac{d}{dx} = 4 \lambda (x-1)+2(x+1)e^y $$ $$ \frac{d}{dy} = 2 \lambda y + (x+1)^2 e^y $$ $$ \frac{d}{d \lambda} = 2(x-1)^2+y^2-3 $$ Now I should solve the system of equations, don't I ? But how? How would you proceed? I thought that I should solve the system for a certain variable, but I'm really not sure...so I'd be thankful if anybody could help! :)

You equate all these derivatives to $0$. So you get: $$4 \lambda (x-1)+2(x+1)e^y=0\tag{1}$$ $$2 \lambda y + (x+1)^2 e^y=0 \tag{2}$$ $$2(x-1)^2+y^2-3=0\tag{3}$$ From $2$, we have $(x+1) e^y=\frac{-2\lambda y}{x+1}$.

Substitute this in $1$ to get $y=x^2-1$.

Substitute this in $3$ to get $x^4=4x\Rightarrow x=0,4^{\frac{1}{3}}$. Using this we get the possible pairs $\left( x,y\right)$ as $\left( 0,-1\right)$ and $\left( 4^{\frac{1}{3}},4^{\frac{2}{3}}-1\right)$.