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I can't see why this has to be the case.

I've computed the vertices, given the constraint inequalities and equations. So, now all that's left is to plug in these vertices into the function to find my maximal value.

Can someone explain why I was essentially done with the problem (finding the max) as soon as I found the vertices?

user287131
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  • Note that the (true) claim here is that the maximum over the convex polygon (of a linear function) occurs at vertex. It possibly can occur at more than one vertex (and again along edges connecting two such vertices). – hardmath Nov 06 '15 at 01:07

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Hint: For any value of $y, z$ chosen, varying $x$ will increase or decrease the function, unless it is on the boundary of the region and you cannot vary $x$ in the direction that increases it. The same argument holds true for other variables, and you end up with one of the vertices as the only choices for any extrema.

This property holds true for all functions which are linear and defined on a convex polygon.

Macavity
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The rest of the Macavity's post.

Let $\Pi$ be a polygon and $f:\Pi\rightarrow \mathbb{R}$ be a function to maximize.

Case 1. $f$ is an affine function. Then $f$ is defined on the convex hull $C$ of $\Pi$ and the sup is reached in a vertex of $C$, that is in a vertex of $\Pi$.

Case 2. $f$ is any convex function. Then the sup is reached again in a vertex of $\Pi$. To see that, it suffices to decompose $\Pi$ into a union of tetrahedra (we are in $\mathbb{R}^3$).