I have already read multiple questions on this topic but none of them provided me a satisfying answer to the following point:
Let be $f: \mathbb{R}^2 \to \mathbb{R}$, where $f{ x \choose y}=x^2+y^2$ and $g :\mathbb{R}^2 \to \mathbb{R}$, with $g{ x \choose y}=2x-y-5$. We now impose the contraint $g{ x \choose y}=0$. This gives us a set of points ${x \choose y}$ which can be described by a curve. If we follow the curve and plug in the values into $f$ we are simply looking for the highest or lowest value. So far so good.
The image above shows the case where both images of $f$ and $g$ intersect. But what if the images of $f$ and $g$ didn't intersect? (where the image of a function is defined by: $im(f):=\{ z \in \mathbb{R}~|~ \exists {x \choose y} \in \mathbb{R}^2$ with $f{x \choose y}=z\}$). Then I would say that there is no solution to the optimization problem. However, in some questions here on MSE they said that this would make no sense but I don't understand why it would make no sense? What is wrong with this picture in my mind?
May be someone can shed light on this issue.
Your image shows the set $\{(x,y,z):g(x,y)=0\}$ and the set $\{(x,y,z):z=f(x,y)\}$. So this has nothing to do with the image of $g$. However it is correct that we are trying to minimize $z$ over the intersection of those two sets, and there would be no solution if the sets did not intersect. This can only occur if the set of $(x,y)$ with $g(x,y)=0$ fails to intersect the domain of $f$.