minimize simple exponential function

Question

Solve the primal and dual formulation of

$$ \text{minimize } f(x)= e^x \\ \text{subject to } x \geq 1 $$

Answer 1

Noting that $f(x)=e^x$ is an increasing function over its domain, the minimum of $f(x)$ for $x \geqslant 1$ occurs at $x=1$, namely, $f(1)=e$.

Answer 2

I want to answer my question. Kindly verify if there are any errors.

The primal solution is $x^* =1$ i.e $f(x^*)=e$. Consider the dual function:

$$\Theta(\lambda)= \min_{x} e^x + \lambda (1-x)$$ This reaches the minimum at $x^* = ln(\lambda)$ where $\lambda > 0$

Now, on substitution, we get, the dual problem to be

$$\max_{\lambda >0}\Theta(\lambda)= 2\lambda - \lambda ln(\lambda) $$

$\nabla \Theta = 0$ gives $\lambda = e$ which has the same objective value as the primal.

Now, for the dual of dual,

$$\Theta(x)= \max_{\lambda} 2\lambda - \lambda ln(\lambda) -x\lambda$$

this has the maximum at $\lambda = e^{1-x}$ which then gives the original primal problem.