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.
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.