Analyze the optimization problem
min $f(x, y) = e^{-x} + e^{-2y}$ over the entire domain of $f$.
Determine if $f$ is convex, concave or neither. If $f$ is neither, what can you say about its possible solutions?
My work:
- The domain of $f$ is $\mathbb{R}^2$.
- If $f(x,y) = e^{-x} + e^{-2y}$, then $\nabla f (x,y) = (-e^{-x}, -2e^{-2y})$ and $H_f = \begin{pmatrix} e^{-x} & 0 \\0 & 4e^{-2y} \end{pmatrix}$.
- $f$ has no critical points since $\nabla f (x,y) \neq (0,0)$ for any $(x,y) \in \mathbb{R}^2$.
- Using a leading principal minor analysis, we get that $D_1 = e^{-x} > 0$ and $D_2 = 4e^{-x-2y} > 0$ for all $(x,y) \in \mathbb{R}^2$, which implies that $H_f$ is positive definite for all $(x,y)\in \mathbb{R}^2$. Thus, $f$ is strictly convex over its domain.
My conclusions:
- This optimization problem has no solutions since $f$ does not achieve a minimum over its domain (no critical points).
- If $f$ was still strictly convex but did have a critical point, that critical point would necessarily be a global minimum.
- If $f$ was neither convex nor concave over its domain and did have one or more critical points, then we would try to and classify the critical points as local maximums, minimums or saddle points using the second derivative test.
Are my conclusions correct in answering the question?