# 3.2a. Solving Linear Programming Problems Graphically

Wouldn't it be nice if we could simply produce and sell infinitely many units of a product and thus make a never-ending amount of money? In business (and in day-to-day living) we know that we cannot simply choose to do something because it would make sense that it would (unreasonably) accomplish our goal. Instead, our hope is to maximize or minimize some quantity, given a set of constraints.

Think about this: you are traveling from Chandler, AZ, to San Diego, CA. Your hope is to get there in as little time as possible, hence aiming to minimize travel time. At the same time, you will be facing more or less traffic on certain stretches of the trip, you will need to stop for gas at least once (unless you are driving a hybrid vehicle), and, if you have kids, you'll definitely need to stop for a restroom break. While we have only mentioned a few, these are all

constraints—things that limit you in your goal to get to your destination in as little time as possible.

## Solving Linear Programming Problems Graphically

A linear programming problem involves constraints that contain inequalities. An

inequality is denoted with familiar symbols, <, >,
$\le$
, and
$\ge$
. Due to difficulties with strict inequalities (< and >), we will only focus on
$\le$
and
$\ge$
.

In order to have a linear programming problem, we must have:

• Inequality constraints
• An objective function, that is, a function whose value we either want to be as large as possible (want to maximize it) or as small as possible (want to minimize it).

### Example 1

An airline offers coach and first-class tickets. For the airline to be profitable, it must sell a minimum of 25 first-class tickets and a minimum of 40 coach tickets. The company makes a profit of $225 for each coach ticket and$200 for each first-class ticket. At most, the plane has a capacity of 150 travelers. How many of each ticket should be sold in order to maximize profits?

#### Solution

The first step is to identify the unknown quantities. We are asked to find the number of each ticket that should be sold. Since there are coach and first-class tickets, we identify those as the unknowns. Let,

x = # of coach tickets

y = # of first-class tickets

Next, we need to identify the objective function. The question often helps us identify the objective function. Since the goal is the maximize profits, our objective is identified.

Profit for coach tickets is $225. If x coach tickets are sold, the total profit for these tickets is 225x. Profit for first-class tickets is$200. Similarly, if

y  first class tickets are sold, the total profit for these tickets is 200y.

The total profit, P, is

P = 225x + 200y

We want to make the value of

as large as possible, provided the constraints are met. In this case, we have the following constraints:

• Sell at least 25 first-class tickets
• Sell at least 40 coach tickets
• No more than 150 tickets can be sold (no more than 150 people can fit on the plane)

We need to quantify these.

• At least 25 first-class tickets means that 25 or more should be sold. That is, y
$\ge$
25
• At least 40 coach tickets means that 40 or more should be sold. That is, x
$\ge$
40
• The sum of first-class and coach tickets should be 150 or fewer. That is x + y
$\le$
150

Thus, the objective function along with the three mathematical constraints is:

Objective Function: P = 225x + 200y

Constraints: y
$\ge$
25;
$\ge$
40; x + y
$\le$
150

We will work to think about these constraints graphically and return to the objective function afterwards. We will thus deal with the following graph:

Note that we are only interested in the first quadrant, since we cannot have negative tickets.

We will first plot each of the inequalities as equations, and then worry about the inequality signs. That is, first plot,

x= 25

y = 40

x + y = 150

The first two equations are horizontal and vertical lines, respectively. To plot x + y= 150, it is preferable to find the horizontal and vertical intercepts.

To find the vertical intercept, we let

x = 0:

y= 150

Giving us the point (0,150)

To find the horizontal intercept, we let

y = 0:

x = 150

Giving us the point (150,0)

Plotting all three equations gives:

Our next task is to take into account the inequalities.

We first ask, when is y
$\ge$
25? Since this is a horizontal line running through a y-value of 25, anything above this line represents a value greater than 25. We denote this by shading above the line:

This tells us that any point in the green shaded region satisfies the constraint that

y
$\ge$
25.

Next, we deal with

x
$\ge$
40. We ask, when is the x-value larger than 40? Values to the left are smaller than 40, so we must shade to the right to get values larger than 40:

The blue area satisfies the second constraint, but since we must satisfy

all constraints, only the region that is green and blue will suffice.

We have one more constraint to consider:

x+ y
$\ge$
150. We have two options, either shade below or shade above. To help us better see that we will, in fact, need to shade below the line, let us consider an ordered pair in both regions. Selecting an ordered pair above the line, such as (64, 130) gives:

64 + 130
$\ge$
150

Which is a false statement since 64 + 130 = 194, a value larger than 150.

According to the graph, the point (64, 65) is one that falls below the graph. Putting this pair in yields the statement:

64 + 65
$\ge$
150

Which is a true statement since 64+65 is 129, a value smaller than 150.

Therefore, we shade below the line:

The region in which the green, blue, and purple shadings intersect satisfies all three constraints. This region is known as the feasible regions, since this set of points is feasible, given all constraints. We can verify that a point chosen in this region satisfies all three constraints. For example, choosing (64, 65) gives:

64
$\ge$
40 TRUE

65
$\ge$
25 TRUE

64 + 65
$\ge$
150 TRUE

This gets us to a great point, but still does not answer the question:

which point maximizes profit? Fortunately, there is a theorem discovered by mathematicians that allows us to answer this question.

First off, we define a new term: a corner point is a point that falls along the corner of a feasible region. In our situation, we have three corner points, shown on the graph as the solid black dots:

The objective function along with the three corner points above forms a bounded linear programming problem. That is, imagine you are looking at three fence posts connected by fencing (black point and lines, respectively). If you were to put your dog in the middle, you could be sure it would not escape (assuming the fence is tall enough). If this is the case, then you have a bounded linear programming problem. If the dog could walk infinitely in any one direction, then the problem is unbounded.

### Fundamental Theorem of Linear Programming

1. If a solution exists to a bounded linear programming problem, then it occurs at one of the corner points.
2. If a feasible region is unbounded, then a maximum value for the objective function does not exist.
3. If a feasible region is unbounded, and the objective function has only positive coefficients, then a minimum value exist

This means we have to choose among three corner points. To verify the "winner," we must see which of these three points maximizes the objective function. To find the corner points as ordered pairs, we must solve three systems of two equations each:

= 40

x + y = 150

= 40

y= 25

#### System 3

y = 25

x+ y = 150

We could decide to solve by using matrix equations, but these equations are all simple enough to solve by hand:

(40) + y = 150

y = 110

Point:(40,110)

Point: (40,25)

#### System 3

x + 25 = 150

x = 125

Point: (125,25)

We test each of these three points in the objective function:

#### Solution

Our goal is to determine the percentage increase for administrative secretaries and faculty, so let

x = percentage increase for secretaries

y = percentage increase for faculty

The college would like to minimize its total expenditures, so the objective function must include the total amount of money outflows. Since the new secretaries will require a total budget of

$28,000 × 8 =$224,000 and the faculty a total budget of $40,000 × 7 =$280,000, the total cost will be the raise percentage for each group, multiplied by the total salaries:

C = 224x + 280y

There is one constraint given, which is that the total raises must be $5,000 or less. That is, 224x + 280y $\le$ 5 Of course, the college does not want to reduce the salaries, so x $\ge$ 0 and y $\ge$ 0. To visualize the situation, we graph the constraint as an equation. To help us find points, we first find the intercepts: Horizontal Intercept: 224(0) + 280 y = 5 y ≈ .018 Vertical Intercept 224x + 280(0) = 5 x ≈ .022 We then plot the points and connect them with a straight line: Since the inequality sign is $\le$ , we shade below the line: This gives us three corner points, as shown above. We test each to verify which of the pairs of percentages gives the minimum cost:  Point Cost (0,0) 224(0) + 280(0) =$0 (0,.018) 224(0) + 280(.018) = $5.04 (.020,0) 224(.020) + 280(0) =$4.48
Clearly, the first option gives the smallest cost; however, this combination of tells us to give a 0% raise to both groups, which, of course, is not practical, since the company's goal was to give a raise to each group.

Why did this happen, and what should we do to fix it? Well, when we think about the constraint of spending $5,000 or less and hoping to make expenditures as small as possible, wouldn't it make sense to say, "don't spend anything!"? This outcome will occur anytime we are minimizing, have constraints with the le inequality sign, and when the origin is included in the feasible region. To fix the problem, the company should make additional specifications, such as, what is the minimum percentage raise to give to each group? Is it desirable for one of the raises to be larger than the other? These are questions the analyst should discuss with human resources and administration. # Practice Problems 1) Solve each of the following linear programming problems. a) Maximize R = 2x + 3y Subject to –2xy $\ge$ –10 x + 3y $\ge$ 6 x $\ge$ 0 y $\ge$ 0 Solution: R=30 at (0,10) B) Minimize T = 3x + y Subject to x + 2y $\ge$ 4 x + 3y $\ge$ 6 x $\ge$ 0 y $\ge$ 0 Solution: R=2 at (0,2) 2) A local school governing board approves a new math education program that is to be implemented at a series of elementary schools within the district. Money for the program will come from two different budgets: public expenditures budget and grade-school initiatives budget. The board is willing to pay at least half of what comes out of the initiatives budget from its public expenditures budget. Since this program is considered an initiative, the government mandates that at least$2,000 comes from the local initiatives budget. Both budgets are partially funded by federal emergency funding. For the public expenditures budget, the percentage is 55% and 23% for the grade-school initiatives budget. In order to properly use emergency funding, the district would like to minimize the use of federal dollars. How much should come from each budget?

Solution: x=amount from publix expenditures; y=amount from grade school initiatives

Minimize: C=0.55x+0.23y

(1/2)x≥y or {(1/2)x-y≥0}

y≥2000

x,y≥0

Solution: C=2660, x=4000, y=2000

3) A public relations director for a homeopathic is seeking to advertise her company's products on two different websites—one is a medical parts supplier and the other is a fitness e-zine (a web-based magazine). The medical parts supplier website receives, on average, about 1,200,000 hits per day per page, while the fitness e-zine receives about 2,000,000 hits per day per page. The daily cost to advertise is $1,100 per advertisement and$1,600 per advertisement, respectively. The director would like at least 15 ads and is able to allocate up to \$50,000 for advertising. At least 3 ads should be placed on each website. How many adds should be placed on each website to maximize the potential number of readers (even if some viewers see the add on different pages of the website)?

x=number on medical website;   y= number on fitness website.

Maximize: R=1200000x+2000000y

Subject to:

x+y≥15

1100x+1600y≤50000

x≥3

y≥3

x,y≥0

 Corner Points R=1200000x+2000000y (3,12) 27,600,000 (12,3) 20,400,000 (3,29.2) 62,000,000 (41.1,3) 55,320,000
Optimal Solution

 (3,29.2) 62,000,000
See more examples in next section.