Let X = # of tubes A and Y = # of tubes B.

Then from the condition you have this system of restrictions


4X + 3Y <= 28  hours   (1)   (counting hours of the carpenter)
4X + 2Y <= 32  hours   (2)   (counting hours of the plumber)


and your goal is to maximize  X + Y, which is the objective function  F(X,Y) = X + Y.


Use the Linear Programming method.


The feasible domain is shown in the plot below.



    


    Plot  4*X + 3*Y = 28 (red);  4*X + 2*Y = 32  (green),  and  X + Y = 9 (blue).



It is the triangle in QI adjacent to X- and Y-coordinate axes and bounded by the red line.


The feasible grid points inside this triangle represent possible solutions (integer solutions) to the system 
of restrictions (inequalities) (1) and (2).


Of them (of these points), we should find (we should select) the one which gives the maximum to the sum X + Y, which is our objective function.


To do it, take a straight line  X+Y = const and move it from the coordinate origin as far as it is possible, in a way it still passes  
through the grid points inside of the feasible triangle.


The blue line showed in the plot gives you THE SOLUTION.

There are TWO solutions (!)


One is       X= 0, Y= 9 with X + Y = 9.                  (on the blue line)

The other is X= 1, Y = 8 with the same sum  X + Y = 9.   (on the blue line, too)


Answer.  The maximum possible number of installed tubes is  X+Y = 9.


          There are TWO possible solutions: one with X= 0 (tubes A)  and Y= 9 (tubes B);
                                           the other X= 1 (tubes A)  and Y= 8 (tubes B)  with the same sum  X + Y = 9.