Let X be the number of type A chairs;

let Y be the number of type B chairs.


The objective function (profit)  is

R(X,Y) = 60X + 84Y.


The restrictions are :

2X + 3Y <= 30       (1)     (restriction on the machine time)   and
5X + 5Y <= 60       (2)     (restriction on the craftsman time).
X >= 0;  Y >= 0.    (3)     (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  2x + 3y = 30  and the green line  5X + 5Y = 60.






Plots y =   (red) and y =  (green)



The method of linear programming says:

    1) Take the vertices of this quadrilateral

        (X1,Y1) = (0,10)   (red line Y-intercept);

        (X2,Y2) = (6,6)   (intersection point of the straight lines Y =  and Y =  );

        (X3,Y3) = (12,0)   (green line X-intercept)


    2) Calculate the objective function at these points

        R(X1,Y1) = 60*0 + 84*10 =  840;

        R(X2,Y2) = 60*6 + 84*6  =  864;

        R(X3,Y3) = 60*12 + 84*0 =  720.


    3) Then select one of these point where the objective function is maximal - In our case this point is (X2,Y2) = (6,6).


    4) This point gives your optimal solution X = 6 chairs of type A and Y = 6 chairs of the type B.


If they follow this optimal solution, their weekly profit will be MAXIMAL, E864.