Linear Programming

Many mathematical models designed to solve problems in business, biology, and economics involve finding the optimum value (maximum or minimum) of a function (the objective function) subject to certain conditions (constraints) expressed as linear inequalities. The inequalities define what we call the feasible region. When only two variables are involved the solution can often be found graphically.

Find the maximum value of z = 3x + 2y subject to the constraints: x ≥ 0 y ≥ 0 x + 2y ≤ 4 x – y ≤ 1

Corner Point Theorem If an optimum value (either a maximum or a minimum) of the objective function exists, it will occur at one or more of the vertices of the feasible region.

The corner points are A = (0,0) B = (1,0) C = (2,1) D =(0,2)

z = 0 at A z = 3 at B z = 8 at C z = 4 at D The maximum value of z is 8, and it occurs when x = 2 and y = 1.

So, now let’s try a REAL problem. An office manager needs to purchase new filing cabinets. Ace cabinets cost $40 each, require 6 square feet of floor space, and hold 24 cubic feet of files. Excello cabinets cost $80, require 8 square feet of floor space, and hold 36 cubic feet of files. She has only $560 in her budget and has only 72 square feet of floor space available. The manager wants to maximize storage capacity given these limitations. How many of each cabinet should she buy? Make a chart to represent the information.

We want to maximize storage capacity, so If x is the number of Aces and y is Excellos z = 24x + 36y Constraints 40x + 80y ≤ 560 6x + 8y ≤ 72 x ≥ 0 y ≥ 0 Now graph and find the vertices of the feasible region.

Which of the points maximizes the function? (0,0)z= 0 (0,7)z = 252 (12,0)z = 288 (8,3)z = 300 How many of each should the manager buy? What will the maximum storage space be?

Comments Problems can also have minimal optimal values. Sometimes no minimum or maximum exists. (A feasible region may be unbounded.) Oftentimes only integer solutions make sense. These problems can be more challenging. If this is the case, our problems will be “contrived” to have integer solutions.

Now you try this… A 4-H member raises only geese and pigs. She does not want to raise any more than 16 animals, and she wants no more than 10 of those to be geese. She spends $5 to raise a goose and $15 to raise a pig. She only has $180 available for the project.The 4-H member wishes to maximize her profits. Each goose produces $6 in profit and each pig $20. How many of each should she raise?

Let x = geese and y = pigs, then z = 6x + 20y x + y ≤ 16 x ≤ 10 5x + 15y ≤ 180 x ≥ 0 y ≥ 0

The vertices of the feasible region are (0,0), (0,12), (6,10), (10,6), and (10,0) Which one produces the maximum profit? (0,12) produces 240 So she should raise 12 pigs to produce $240 profit. Did you “see” this answer right away? How can she maximize her profit by only raising 12 animals?