To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top

Linear Programming Calculator
Top
Linear programming deals with the optimization (maximization or minimization) of linear functions subject to linear constraints. Linear Programming Calculator is a tool which makes calculations easy and fun. Try our Free Linear Programming Calculator and get your problems solved instantly.

You can see a default objective function along with the linear inequalities and set of constraints given below. The calculator will solve the inequalities and find the coordinates and mark the feasible region. And finally find the value of objective function when on clicking "Submit".
 

Steps For Linear Programming Calculator

Back to Top
Step 1 :  

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.



Step 2 :  

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



Step 3 :  

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).



Example problems for Linear Programming Calculator

Back to Top
  1. Solve the following linear program

    maximise 5x1 + 6x2

    subject to

    x1 + x2

    x1 - x2 >= 3

    5x1 + 4x2

    x1 >= 0

    x2 >= 0


    Step 1 :  

    It is plain that the maximum occurs at the intersection of

    5x1 + 4x2 = 35 and

    x1 - x2 = 3



    Step 2 :  

    Solving simultaneously, rather than by reading values off the graph, we have that

    5(3 + x2) + 4x2 = 35

    i.e. 15 + 9x2 = 35



    Step 3 :  

    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



    Answer  :  

    39.444



  2. 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 .


    Step 1 :  

    Variables

    Let

    xT = number of tables made per week

    xC = number of chairs made per week



    Step 2 :  

    Constraints
    total work time

    6xT + 3xC
    customer demand

    xC >= 3xT
    storage space

    (xC/4) + xT
    all variables >= 0



    Step 3 :  

    Objective

    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



    Answer  :  

    xC = 10.667, xT = 1.333 and the profit = £146.667



*AP and SAT are registered trademarks of the College Board.