Problem 45 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Travel has decided to advertise in the Sunday editions of two major newspapers in town. These advertisements are directed at three groups of potential customers. Each advertisement in newspaper I is seen by 70,000 group-A customers, 40,000 group-B customers, and 20,000 group-C customers. Each advertisement in newspaper II is seen by 10,000 group-A, 20,000 group-B, and 40,000 group-C customers. Each advertisement in newspaper I costs $$\$ 1000$$, and each advertisement in newspaper II costs $$\$ 800$$. Everest would like their advertisements to be read by at least 2 million people from group A. \(1.4\) million people from group \(\mathrm{B}\), and 1 million people from group C. How many advertisements should Everest place in each newspaper to achieve its advertising goals at a minimum cost? What is the minimum cost?

Step-by-Step Solution

Answer

Everest should place no advertisements in newspaper I and 50 advertisements in newspaper II to achieve their advertising goals at a minimum cost of $$\$ 40,000$$.

Step 1: Define Variables

Let the number of ads in newspaper I be \(x\), and the number of ads in newspaper II be \(y\).

Step 2: Objective Function

We need to minimize the advertising cost. The total cost of advertising in newspaper I is \(1000x\), and for newspaper II, it is \(800y\). The objective function to minimize is: \[C(x, y) = 1000x + 800y\]

Step 3: Define Constraints

Now, we need to define the constraints based on the target audience for each group: 1. For group A: \(70,000x + 10,000y \geq 2,000,000\) 2. For group B: \(40,000x + 20,000y \geq 1,400,000\) 3. For group C: \(20,000x + 40,000y \geq 1,000,000\) Also, the number of advertisements cannot be negative: 4. \(x \geq 0\) 5. \(y \geq 0\)

Step 4: Convert Constraints to Standard Form

Dividing the constraints by the respective thousands, and rearrange to the standard form: 1. \(7x + y \geq 200\) 2. \(2x + y \geq 70\) 3. \(x + 2y \geq 50\) 4. \(x \geq 0\) 5. \(y \geq 0\)

Step 5: Solve Linear Programming Problem

We will now solve the Linear Programming problem using the graphical method, which consists of plotting the constraint equations on a Cartesian plane, finding the feasible region, and identifying the vertices. Finally, we will calculate the cost for each vertex and find the minimum cost vertex. The intersection points (vertices) of the feasible region are: 1. \((A_1) = (50, 0)\) 2. \((A_2) = (20, 30)\) 3. \((A_3) = (0, 50)\) Compute the cost for each vertex: \(C(A_1) = 1000 \cdot 50 + 800 \cdot 0 = 50,000\) \(C(A_2) = 1000 \cdot 20 + 800 \cdot 30 = 44,000\) \(C(A_3) = 1000 \cdot 0 + 800 \cdot 50 = 40,000\) The minimum cost is at point \(A_3\) with a cost of $$\$ 40,000$$.

Step 6: Conclusion and Answer

Everest should place no advertisements in newspaper I and 50 advertisements in newspaper II to achieve the minimum cost of advertising. The minimum cost for advertising is $$\$ 40,000$$.

Key Concepts

Objective Function

In linear programming, the objective function is the mathematical expression that needs to be optimized, either maximized or minimized, depending on the problem at hand. For example, in the advertisement problem involving newspapers, Everest wants to minimize costs.
The objective function here mathematically represents the total cost of placing advertisements in two newspapers. Each ad in newspaper I costs \\(1000, and in newspaper II, it costs \\)800. If we let \(x\) and \(y\) be the number of ads in newspapers I and II, respectively, the objective function is:

• \[ C(x, y) = 1000x + 800y \]

This equation adds the cost of all ads in both newspapers. Everest's goal is to find the values of \(x\) and \(y\) that make this cost as low as possible while still meeting customer reach requirements.

Constraints

Constraints are conditions that must be met in any solution of a linear programming problem. They restrict the feasible solutions to a specific region in the graph.
In this advertising problem, the constraints arise from the need to reach a certain number of customers in three different groups. For instance:

• Group A needs to see at least 2 million ads, leading to the constraint \(70,000x + 10,000y \geq 2,000,000\).
• Group B requires 1.4 million views, providing the constraint \(40,000x + 20,000y \geq 1,400,000\).
• Group C demands 1 million views, resulting in the constraint \(20,000x + 40,000y \geq 1,000,000\).
• Non-negativity constraints: \(x \geq 0\) and \(y \geq 0\), since negative ads aren’t possible.

These constraints are essential because they help ensure that the solution is practical and meets necessary real-world conditions.

Feasible Region

The feasible region is the set of all possible points that satisfy the constraints of the linear programming problem. Graphically, it's the area on a graph where all the inequality conditions overlap.
In our advertising example, each constraint lines represents a boundary on the graph. By plotting these on the Cartesian plane:

• \(7x + y \geq 200\)
• \(2x + y \geq 70\)
• \(x + 2y \geq 50\)

The overlapping area, where all these conditions are true, represents our feasible region.
This region is critical because it contains all possible combinations of \(x\) and \(y\) that satisfy the advertising goals for different customer groups.

Minimum Cost Solution

The minimum cost solution in linear programming is the point in the feasible region that yields the lowest value of the objective function. In the problem, we minimize the cost of advertising which is represented by the objective function.
To find this solution, we evaluate the cost function at each vertex of the feasible region. The vertices are points where the constraint lines intersect.
For example, the vertices found were:

• (50, 0) with a cost of \\(50,000
• (20, 30) with a cost of \\)44,000
• (0, 50) with a cost of \\(40,000

After calculating the costs at these points, the minimum cost was discovered at the vertex (0, 50), where Everest places 0 ads in newspaper I and 50 ads in newspaper II, resulting in the lowest total cost of \\)40,000.
This solution is optimal because it meets all constraints while minimizing costs, achieving Everest's advertising goals efficiently.