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 =
Plots y =  (red) and 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 =
 (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 =
 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.
 );
        (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.
Solved.
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For many other similar solved problems see the lesson
    - Solving minimax problems by the Linear Programming method 
in this site.
Learn from there the technique and the methodology on how to solve minimax problems using the Linear Programming method - once and for all.