Lesson STEVE WILSON MATH 131 LINEAR PROGRAMMING PROBLEM NUMBER 1 SOLUTION

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STEVE WILSON MATH 131 LINEAR PROGRAMMING PROBLEM NUMBER 1

A farmer has 10 acres to plant in wheat and rye.  He has to plant at least 7 acres.  However, he has only $1200 to spend and each acre of wheat costs $200 to plant and each acre of rye costs $100 to plant.  Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of wheat and 2 hours to plant an acre of rye.  If the profit is $500 per acre of wheat and $300 per acre of rye how many acres of each should be planted to maximize profit?

You need to find the objective equation and the constraint equations and then you need to graph the constraint equations to find the area of compatibility within which your answer will lie.  The maximum and minimum values on this graph will always be at the corners of the area of compatibility (the corners of the area of compatibility are at the intersection points of the lines on the graph).

First thing to do is find the objective equation.  This is the equation that contains the variables that need to be maximized or minimized.  This is sometimes also called the objective function.   A function is another term for an equation where there is only 1 value of the dependent variable for each value of the independent variable.   y is usually the dependent variable, but not always.  x is usually the independent variable, but not always.

Next thing you need to do is find the constraint equations.  These are the equations that limit the range of each of the constraint variables.  These equations will be graphed so you can find the area of compatibility.  The area of compatibility is the area on the graph that satisfies all the constraints of all the equations simultaneously.

First figure out what variables you want to use and what they will represent.

x will represent the number of acres of wheat that will be planted.
y will represent the number of acres of rye that will be planted.

The objective function will be:

profit = 500x + 300y 
This is because each acre of wheat provides 500 dollars of profit and each acre of rye provides 300 dollars of profit.

the constraint equations will be:

x + y >= 7 
This is because the total number of acres planted has to be greater than or equal to 7.

x + y <= 10 
This is because the total number of acres planted has to be less than or equal to 10.

x + 2y <= 12 
This is because the total number of hours available for planting is 12 and it takes twice as many hours to plant an acre of rye than an acre of wheat.

200x + 100y <= 1200 
This is because it takes 200 dollars to plant an acre of wheat and 100 dollars to plant an acre of rye and the total amount of dollars available is equal to 1200.

x >= 0
y >= 0
This is because the number of acres can never be negative.

You need to graph the constraint equations.  In order to do that on most graphing software, you need to solve for y in each of the constraint equations.

x + y >= 7 becomes y >= (7-x)

x + y <= 10 becomes y <= (10-x)

x + 2y <= 12 becomes y <= (12 - x) / 2

200x + 100y <= 1200 becomes y <= (1200 - 200x) / 100

You now need to graph these equations.  When you graph inequality equations, you need to graph the equality portion of the equations to create the lines and then shade the area that is compatible with the inequality portion of the equations.

In the following graph, arrows are used to show the area of compatibility with the inequality portion of the equations.  These arrows connect to the line of the equality portion of the equation and show the direction of the area of compatibility of the inequality portion of the equation.

The graph is shown below:

As you can see:

The area of compatibility for y >= (7-x) is the area on the line of y = (7-x) and above it.
The area of compatibility for y <= (10-x) is the area on the line of y = (10-x) and below it.
The area of compatibility for y <= (12-x)/2 is the area on the line of y = (12-x)/2 and below it.
The area of compatibility for y <= (1200 - 200x)/100 is the area on the line y = (1200 - 200x)/100 and below it.

The area of compatibility that meets all the constraints of all the equations simultaneously is the area within the triangle marked ABC.   This is the area that meets all of the constraints of all of the equations simultaneously.

Your answer will lie on the corners of this area which is the intersection points of the lines that bound this area.

Those points are:

A(2,5)
B(4,4)
C(5,2)

You need to evaluate the objective equation at each of these points.

The objective equation is profit = 500x + 300y.

At point A, x = 2 and y = 5 and the objective function is profit = 500(2) + 300(5) which becomes profit = 1000 + 1500 which becomes profit = 2500.

At point B, x = 4 and y = 4 and the objective function is profit = 500(4) + 300(4) which becomes profit = 2000 + 1200 which becomes profit = 3200.

At point C, x = 5 and y = 2 and the objective function is profit = 500(5) + 300(2) which becomes profit = 2500 + 600 which becomes profit = 3100.

All the constraints have been met and your profit is maximized when x = 4 and y = 4 which is at point B.

The intersection points themselves are found by several methods, the easiest being to use a graphing calculator that tells you what the intersection points are.

I use a TI-84 at home (TI-83 will do the same for you) and I use the Desmos Graphing Calculator online.

You can also do it manually by following the rules for solving equations simultaneously.

Here's a reference on linear equations that includes references and instructions on how to solve linear equations simultaneously.

<a href = "http://www.algebra.com/algebra/homework/Linear-equations/theo-20090927.lesson" target = "_blank">http://www.algebra.com/algebra/homework/Linear-equations/theo-20090927.lesson</a>