How to merge two finite state automata?

Say I have two deterministic finite state automata represented by the following transition diagrams:

FSA for keyword IF: IF

  ___        ___         _
 /   \  I   /   \  F   // \\
>| 0 |----->| 1 |----->||2||
 \___/      \___/      \\_//

FSA for an ID: [A-Z][A-Z0-9]*

            ------------
  ___       | _    LET |
 /   \ LET  // \\<------
>| 0 |----->||1||
 \___/      \\_//<------
            |      NUM |
            ------------

What algorithm may I use to combine them into a single deterministic finite state automata with three final states, represented by the following transition diagram:

            -----------------------
            | LETTER BUT F OR NUM |   --------
  ___       | _          _   LET  v _ |  LET |
 /   \  I   // \\  F   // \\----->// \\<------
>| 0 |----->||1||----->||2||      ||3||<--------
 \___/      \\_//      \\_//----->\\_//<------ |
   |                         NUM  |      NUM | |
   | ANY LETTER OTHER THAN I      ------------ |
   ---------------------------------------------

1: ID
2: IF (IT'S ALSO AN ID, BUT THE KEYWORD IF HAS A HIGHER PRECEDENCE)
3: ID

The textbooks usually gives the automaton C such that L(C) = L(A) U L(B) by applying de-morgan on it, L(C) = (L(A)^C [intersection] L(B)^C)^C.
The intersection is done by building the Cartesian product automaton, and the negation is simply switching the accepting states.
Building the union automaton can also be done directly: Build the Cartesian product automaton, and a final state is a state (a,b) such that a is a final state in the automaton of A OR b is a final state in the automaton of B

An alternative is building a Non-Deterministic Final Automaton (NFA) by simply creating a new state, and add an epsilon path for both start(A) and start(B), and use the standard algorithm for eliminating non-determinism from an automaton.

The problem - this automaton will not be minimal (far from it probably). You can try and use this algorithm on the resulting automaton in order to minimze it.