How to Use the Graphical Method for Linear Optimization

Here, you’ll learn how to use the substitution method for linear optimization.

Rule

The Substitution Method

1.
Set up inequalities based on the text of the problem you’re given. We’ll be talking about numbers of units, so it’s natural to say that x 0 and y 0, because it’s impossible to produce a negative number of units.
2.
Solve the inequalities for y, so that they look like functions y(x).
3.
Draw the graphs of these functions in a coordinate system. Because x 0 and y 0, you only need to think about the first quadrant.
4.
Mark the area that is surrounded by the graphs!
5.
Call the different corners of the area A, B, and so on.
6.
Find the coordinates of the points A, B, and so on.
7.
Enter the coordinates of A, B, and so on into the objective function (see below) and see which point yields the largest answer. That is the optimal point.

Theory

The Objective Function

The objective function is often a profit or income function that is determined by A, which is the cost of an item that has sold x units, and B, the cost of an item that has sold y units.

Z(x,y) = Ax + By

Example 1

A factory that produces branded merchandise will make a shirt and a skirt for Tom Ford. Both pieces have to be made of cashmere and silk.

To make a shirt, you need two lengths of cashmere and one length of silk.

To make a skirt, you need one length of cashmere and three lengths of silk.

The factory has 200 lengths of cashmere and 300 lengths of silk available. A shirt is sold for $295, while a skirt is sold for $450. How many shirts and skirts does the factory have to produce to maximize its income, and what is this income?

Item 1

First, you set up the constraints from the text as inequalities. Let x be the number of shirts produced and y be the number of skirts produced. The cashmere and the silk has to be distributed between shirts and skirts. That gives you these inequalities:

  • The number of shirts you produce can be no shirts, or more. This means that

    x 0
  • The number of skirts you produce can be no skirts, or more. That gives you

    y 0
  • Now you set up an inequality for the available cashmere. You need two lengths for a shirt and one length for a skirt. The roll has a total of 200 lengths. That means the inequality is as follows:

    2x + y 200
  • Next, you write out an inequality for the available silk. You need one length for a shirt and three lengths for a skirt. The roll has 300 lengths of silk. That results in:

    x + 3y 300

You end up with this system of inequalities:

x 0 (1) y 0 (2) 2x + y 200 (3) x + 3y 300 (4)

Item 2

Now you need solve the inequalities with respect to y. First, inequality (3):

2x + y 200 y 200 2x

Inequality (4):

x + 3y 300 3y 300 x| ÷ 3 y 100 1 3x

You don’t have to do anything about the other two inequalities.

Items 3 and 4

Graph the two inequalities in a coordinate system and mark the area where they overlap. You know that you only need to graph in the the first quadrant, because both x and y are greater than or equal to 0. That should give you this figure:

Example of linear optimization using substitution method

Example of linear optimization using substitution method

Items 5 and 6

Find the optimal point from this figure. You know that it is one of the points within the marked area where the graphs intersect, or where the graphs intersect the axes. When you use the substitution method, you have to find all the points of intersection. You can see from the graph that

  • Point A is where x + 3y = 300 intersects the y-axis. There, x = 0. That gives you

    0 + 3y = 300 3y = 300 y = 100

    This is the point A = (0, 100), which represents producing 0 shirts and 100 skirts.

  • Point B is where x + 3y = 300 intersects 2x + y = 200. This gives you a system equations to solve. You previously rewrote inequalities (3) and (4) so that y was on the left. If you considered them as equations instead, you would get the same answer, except you exchange for an equal sign. That means you have

    y = 200 2x y = 100 1 3x

    Now you can set the expressions for y equal to each other and get

    200 2x = 100 1 3x 200 100 = 2x 1 3x 100 = 2x 1 3x | 3 300 = 6x x 300 = 5x | ÷ 5 60 = x

    Insert this into y = 200 2x, as that is the simplest expression. You get that

    y = 200 2 60 = 200 120 = 80

    This gives you the point B = (60, 80), which represents producing 60 shirts and 80 skirts.

  • Point C is where 2x + y = 200 intersects the x-axis. There, y = 0. That gives you

    2x + 0 = 200 2x = 200| ÷ 2 x = 100

    This gives you the point C = (100, 0), which represents the production of 100 shirts and 0 skirts.

  • Point D = (0, 0) represents producing 0 shirts and 0 skirts.

The different points above show different production combinations for shirts and skirts.

Item 7

You calculate the income of the factory by putting the x-values and y-values you found in the previous point into the income function Z(x,y) and see which point gives the largest answer.

You know that a shirt costs $295 and that a skirt costs $450. Insert those for A and B in the Z(x,y) formula (Z(x,y) = Ax + By), which gives you

Z(x,y) = 295x + 450y

You now calculate the income from the various productions:

The income Z for the point A = (0, 100) is

Z(0, 100) = 295 0 + 450 100 = 45000

The income Z for the point B = (60, 80) is

Z(60, 80) = 295 60 + 450 80 = 17700 + 36000 = 53700

The income Z for the point C = (100, 0) is

Z(100, 0) = 295 100 + 450 0 = 29500

The income Z for the point D = (0, 0) is

Z(0, 0) = 295 0 + 450 0, = 0

That means the factory earns $53700 by producing 60 shirts and 80 skirts. This is the optimal production, because the other production volumes resulted in lower earnings.

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