Identify the unknowns in the given LPP. Denote then by x and y then formulate the objective function in terms of x and y. be sure whether it is to be maximized or minimized.
Translate all the constraints in the form of linear inequations.
Solve these inequations simultaneously. Mark the common area by shaded region which will be the feasible region and now find the coordinates of all the vertices of the feasible region
Find the value of the objective function at each vertex of the feasible region also find the values of x and y for which the objective function z = ax + by has maximum or minimum value (as the case may be).
Solve the following linear program
maximise 5x1 + 6x2
x1 + x2
x1 - x2 >= 3
5x1 + 4x2
x1 >= 0
x2 >= 0
It is plain that the maximum occurs at the intersection of
5x1 + 4x2 = 35 and
x1 - x2 = 3
Solving simultaneously, rather than by reading values off the graph, we have that
5(3 + x2) + 4x2 = 35
i.e. 15 + 9x2 = 35
i.e. x2 = (20/9) = 2.222 and
x1 = 3 + x2 = (47/9) = 5.222
The maximum value is 5(47/9) + 6(20/9) = (355/9) = 39.444
A carpenter makes tables and chairs. Each table can be sold for a profit of £30 and each chair for a profit of £10. The carpenter can afford to spend up to 40 hours per week working and takes six hours to make a table and three hours to make a chair. Customer demand requires that he makes at least three times as many chairs as tables. Tables take up four times as much storage space as chairs and there is room for at most four tables each week.
Formulate this problem as a linear programming problem .
xT = number of tables made per week
xC = number of chairs made per week
total work time
6xT + 3xC
xC >= 3xT
(xC/4) + xT
all variables >= 0
maximise 30xT + 10xC
The graphical representation of the problem is given below and from that we have that the solution lies at the intersection of
(xC/4) + xT = 4 and 6xT + 3xC = 40
Solving these two equations simultaneously we get xC = 10.667, xT = 1.333 and the profit = £146.667
xC = 10.667, xT = 1.333 and the profit = £146.667