TOPIC 9: Linear Programming

Linear Programming

Linear programming is the process of taking various linear inequalities relating to some situation, and finding the "best" value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions.

Formula :

 Y > Mx + C

M = gradient

C = y-intercept

Example 1:

Find the maximal and minimal value of z = 3x + 4y subject to the following constraints:

The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula "z = 3x + 4y" is the optimization equation. I need to find the (x, y) corner points of the feasibility region that return the largest and smallest values of z.

My first step is to solve each inequality for the more-easily graphed equivalent forms:

It's easy to graph the line:

to find the corner points -- which aren't always clear from the graph -- I'll pair the lines equations and solve:



y = –( 1/2 )x + 7y = 3x	y = –( 1/2 )x + 7y = x – 2	y = 3x y = x – 2
–( 1/2 )x + 7 = 3x–x + 14 = 6x14 = 7x 2 = x y = 3(2) = 6	–( 1/2 )x + 7 = x – 2 –x + 14 = 2x – 4 18 = 3x 6 = x y = (6) – 2 = 4	3x = x – 2 2x = –2x = –1 y = 3(–1) = –3
corner point at (2, 6)	corner point at (6, 4)	corner pt. at (–1, –3)


So the corner points are (2, 6), (6, 4), and (–1, –3).
Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, I only need to plug these three points into "z = 3x + 4y".

(2, 6):      z = 3(2)   + 4(6)   =   6 + 24 =   30 
(6, 4):      z = 3(6)   + 4(4)   = 18 + 16 =   34 
(–1, –3):  z = 3(–1) + 4(–3) = –3 – 12 = –15
Then the maximum of z = 34 occurs at (6, 4),
and the minimum of z = –15 occurs at (–1, –3).


Example 2: 

Given the following constraints, maximize and minimize the value of z = –0.4x + 3.2y.

First I'll solve the fourth and fifth constraints for easier graphing:

The feasibility region looks like this:

From the graph, I can see which lines cross to form the corners, so I know which lines to pair up in order to verify the coordinates. I'll start at the "top" of the shaded area and work my way clockwise around the edges:

y = –x + 7y = x + 5	y = –x + 7x = 5	x = 5y = 0
–x + 7 = x + 5 2 = 2x 1 = x y = (1) + 5 = 6	y = –(5) + 7 = 2	[nothing to do]
corner at (1, 6)	corner at (5, 2)	corner at (5, 0)

y = 0y = –( 1/2 )x + 2	y = –( 1/2 )x + 2x = 0	x = 0y = x + 5
–( 1/2 )x + 2 = 0 2 = (1/2)x 4 = x	y = –( 1/2 )(0) + 2 y = 0 + 2 y = 2	y = (0) + 5 = 5
corner at (4, 0)	corner at (0, 2)	corner at (0, 5)

Now I'll plug each corner point into the optimization equation, z = –0.4x + 3.2y:

(1, 6):  z = –0.4(1) + 3.2(6) = –0.4 + 19.2 = 18.8 
(5, 2):  z = –0.4(5) + 3.2(2) = –2.0 + 6.4   =   4.4 
(5, 0):  z = –0.4(5) + 3.2(0) = –2.0 + 0.0   = –2.0 
(4, 0):  z = –0.4(4) + 3.2(0) = –1.6 + 0.0   = –1.6 
(0, 2):  z = –0.4(0) + 3.2(2) = –0.0 + 6.4   =   6.4 
(0, 5):  z = –0.4(0) + 3.2(5) = –0.0 + 16.0 = 16.0


Then the maximum is 18.8 at (1, 6) and the minimum is –2 at (5, 0).

Example 3:

A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.
If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?

The question asks for the optimal number of calculators, so my variables will stand for that:

x: number of scientific calculators producedy: number of graphing calculators produced

Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > –x + 200. The profit relation will be my optimization equation: P = –2x + 5y. So the entire system is:

P = –2x + 5y, subject to: 
100 < x < 200 
80 <  y < 170 
y > –x + 200 
The feasibility region graphs as:   



                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     
When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators".


 
Questions for linear programming.

1.  A gold processor has two sources of gold ore, source A and source B. In order to kep his plant running, at least three tons of ore must be processed each day. Ore from source A costs $20 per ton to process, and ore from source B costs $10 per ton to process. Costs must be kept to less than $80 per day. Moreover, Federal Regulations require that the amount of ore from source B cannot exceed twice the amount of ore from source A. If ore from source A yields 2 oz. of gold per ton, and ore from source B yields 3 oz. of gold per ton, how many tons of ore from both sources must be processed each day to maximize the amount of gold extracted subject to the above constraints

       2.   Find the equation of the line Passing through ( 3, -2) and (-1,4)


       3.   if (a, 2 ) lies on the line 3y = x+1, find a