I want to maximise the score of the following table, choosing one item from each column/row, so no two items are on the same row or column. Score to maximise is just adding all the choices together.
$$*\quad A\quad B\quad C\quad D\quad E$$ $$\alpha\quad 16\quad 16\quad 18\quad 18\quad 18$$ $$\beta\quad 20\quad 18\quad 16\quad 12\quad 10$$ $$\gamma\quad 20\quad 18\quad 18\quad 16\quad 16$$ $$\delta\quad 18\quad 18\quad 16\quad 16\quad 8$$ $$\epsilon\quad 10\quad 12\quad 14\quad 14\quad 14$$
Example: ($^{C}$ means chosen)
$$*\quad A\quad B\quad C\quad D\quad E$$ $$\alpha\quad 16^{C}\quad 16\quad 18\quad 18\quad 18$$ $$\beta\quad 20\quad 18\quad 16\quad 12\quad 10^{C}$$ $$\gamma\quad 20\quad 18^{C}\quad 18\quad 16\quad 16$$ $$\delta\quad 18\quad 18\quad 16^{C}\quad 16\quad 8$$ $$\epsilon\quad 10\quad 12\quad 14\quad 14^{C}\quad 14$$
Gives a score of $16+10+18+16+14=74$
Now there are a few ways to do this, but can firstly, someone actually tell me if $88$ really is the best result, and secondly show me how to do it via graph theory. I think I have done it by graph theory matching myself, and will put up my solution as an answer below.
Yes, $88$ is the maximum.
First, a score of $88$ can be attained, e.g., by choosing $\alpha E,\beta A,\gamma C,\delta B,\epsilon D$, for a score of $18+20+18+18+14=88$ (I think there are $6$ ways to get a score of $88$ but that's not important.)
To beat $88$, you would need to make at least $90$, since (for some silly reason) all the entries are even numbers. If you were allowed to take the biggest element from each row, you would get $18+20+20+18+14=90$. But you can't do that, because the two $20$s are in the same column. So $88$ is the maximum, like I said.
Alternatively, you could just run through the $5!=120$ possible selections by hand in a few minutes.