Let X be the number of tables, and

let Y be the number of chairs.


The objective function (profit)  is

R(X,Y) = 100X + 60Y.   (1)


The restrictions are :

4X +   3Y <= 305       (2)     (restriction on the assembly time)   and
2X + 1.5Y <= 355       (3)     (restriction on the finishing time).
X >= 0;  Y >= 0.       (4)     (non-negativity).


You need to maximize the objective function (profit) under given restrictions.


The feasible domain is shown below.


It is  a quadrilateral in the first quadrant  (X >= 0,  Y >= 0)  restricted 
by the red line  4X +   3Y = 305  and the green line  2X + 1.5Y = 355.



    


    Plots  4X + 3Y = 305  (red) and 2X + 1.5Y = 355 (green)


Now I need to make couple of important notices.


1.  The potential solutions are the points in this quadrilateral with integer coordinates X and Y.


    Therefore, I show in the plot the grid of points with integer coordinates. But, due to technical restrictions, 
    my grid is with the step 15 in both axes.

    Actually, I'd should to show you the grid with the step 1 in both axes, but it would be not impossible to see on the screen.

    Therefore my grid is with the step 15, and I ask you to stretch your imagination and to think that it is the grid with the step 1. 


2.  Next, from the plot you can see that working constrain is, actually, the green line.

    The other, the red line, only constrains the set of possible solution, but does not work as real constrain in the search 
    of the maximal profit.


3.  The blue line in the plot is the line  100X + 60Y = const.

    The left side represent the profit function, but it is not a real profit function: it is some its  "ghost"/"phantom" - 
    - its section by the plane P = const.

    Actually, we should move this blue line PARALLEL TO ITSELF from the "far outside" area of quadrant QI  closer 
    and closer to the origin until this line touches for the first time a grid point in the feasible domain.

    Then the value of the profit function at this point will be the maximal profit, and the point itself with its coordinates
    will give us the solution in terms of tables and chairs.


4.  From the plot, it is clear that it will happen at high values of X close to x-intercept of the green line  x=  = 76.25.

    The simplest way to find this value of X is to express Y =  from (2)  and then find its 
     integer solution in X and Y in that domain.

    It can be done MOMENTARILY (I used Excel, and it gave me integer solution  (X,Y) = (74,3) ).


Thus the optimal solution is the point X = 74 tables and Y = 3 chairs.


If they follow this optimal solution, their weekly profit will be MAXIMAL, P(X,Y) = 100*X + 60*Y = 100*74 + 60*3 = 7580.