minimum $f(x,y) = x + y$ over all $(x,y) \in \mathbb{R}^2 s.t. e^x - 2x + e^y - 2y = 2$

Question

Lets call the set with all the points $(x,y) \in \mathbb{R}^2 s.t. e^x - 2x + e^y - 2y = 2$, M. There exist a function $g(x,y) := e^x - 2x + e^y - 2y - 2$ s.t $g^{-1}(0) = M$. With the Lagrange-multiplicator-method we find that $(x,y) \in M$ is a critical point iff $Df(x,y) = \lambda Dg(x,y)$, this implies that $x$ must equal $y$. I am now stuck on the math part. I don't know how to proof that $M$ is bounded s.t a minimum exists and I don't know how to solve $2e^x -4x = 2$. By drawing a picture on Wolphram Alpha I concluded that $M$ is bounded and has a minimum with respect to $f$ in the point $(0,0)$ but was wondering how to do it without a computer.

Answer 1

This problem can be solved with tangency. We have a smooth curve

$$ c(x,y) = e^x-2x+e^y-2y-2=0 $$

and we ask for the $\lambda$ value in

$$ f(x,y,\lambda) = x+y - \lambda=0 $$

such that $c(x,y)$ and $f(x,y,\lambda^*)$ are tangent.

After substitution we have

$$ e^x-2x+e^{\lambda}e^{-x}-2(\lambda-x)-2=0 $$

then

$$ e^x = 1+\lambda\pm\sqrt{(1+\lambda)^2-e^{\lambda}} $$

but at tangency, $x$ should be unique then

$$ (1+\lambda)^2-e^{\lambda}=0\ \ \ \ \ (*) $$

and using the Lambert function $W$ we have the results of this equation

$$ \lambda = \left\{0,-2 W\left(\frac{1}{2 \sqrt{e}}\right)-1,-2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)-1\right\} $$

or

$$ \lambda = \{0,-1.47767,2.51286\} $$

Follows a plot showing the results for $\lambda = \{0,2.51286\}$.

NOTE

Regarding the equation $(*)$, it is obvious that $\lambda=0$ is a solution: now making $r = \lambda+1$,

$$ r^2= e^{-1}e^r\Rightarrow \pm r = e^{-\frac 12}e^{\frac r2} $$

now following with $r = e^{-\frac 12}e^{\frac r2}$ we have

$$ \frac r2e^{-\frac r2}= \frac{1}{2\sqrt{e}} $$

etc.