Problem 62 - Precalculus with Limits

Question

Advertising A health insurance company advertises on television, on radio, and in the local newspaper. The marketing department has an advertising budget of \(\$ 42,000\) per month. A television ad costs \(\$ 1000 ,\) a radio ad costs \(\$ 200 ,\) and a newspaper ad costs \(\$ 500 .\) The department wants to run 60 ads per month and have as many television ads as radio and newspaper ads combined. How many of each type of ad can the department run each month?

Step-by-Step Solution

Answer

The department can run 30 television ads, 18 radio ads and 12 newspaper ads each month.

Step 1: Define variables

Let's first define the variables: let \( x \) be the number of television ads, \( y \) be the number of radio ads, and \( z \) be the number of newspaper ads.

Step 2: Setup the equations

The problem gives us three important pieces of information that allow us to set up three equations. The first is that the total number of ads is 60 which leads to our first equation: \(x + y + z = 60\). The second is that the total advertising budget is $42,000. This information leads to our second equation: \(1000x + 200y + 500z = 42000\). And finally, the number of television ads is as many as radio and newspaper ads combined, giving us our third equation: \(x = y + z\).

Step 3: Substitute the third equation into the other two

It is easier to solve the system when we have two equations, two variables. So we substitute \(x\) for \(y + z\) into the first two equations. The first equation becomes \(y + z + y + z = 60\), which simplifies to \(2y + 2z = 60\), and simplifies further to \(y + z = 30\). The second equation becomes \(1000(y + z) + 200y + 500z = 42000\), which simplifies to \(y + 2z = 42\), and again further simplifies to \(y + 2z = 14\). This leaves us with two equations and two variables which are easier to solve.

Step 4: Solve for y and z

Subtract the equation \(y + z = 30\) from the equation \(y + 2z = 42\). This results in \(z = 12\). Substitute this into the equation \(y + z = 30\) to find \(y = 18\).

Step 5: Solve for x

Substitute \(y = 18\) and \(z = 12\) into the third original equation \(x = y + z\) to find \(x = 18 + 12 = 30\).

Step 6: Verify the result

Finally, verify whether these solutions satisfy the three original equations. If not, seek the error in the calculation. If they do, write out the results as a statement.

Key Concepts

Systems of Equations

Linear programming often involves solving systems of equations, especially when multiple relationships need to be balanced simultaneously. In the context of the advertising budget problem, we have three conditions that translate into equations:

• Number of ads: \( x + y + z = 60 \)
• Budget constraint: \( 1000x + 200y + 500z = 42000 \)
• Ad type condition: \( x = y + z \)

These equations together form a system of equations. Solving them means finding values for the variables \( x, y, \) and \( z \) that fit all the conditions simultaneously. This is crucial in ensuring that all requirements are satisfied, such as meeting both the budget and the ad count without exceeding the limits. Systems of equations provide a powerful framework for tackling such problems with multiple constraints.

Budget Constraints

Budget constraints are a common component of linear programming problems, representing the limits within which choices must be made. For our problem, the health insurance company must allocate its monthly budget of \( \\(42,000 \) across different advertising platforms. The equation \( 1000x + 200y + 500z = 42000 \) reflects this constraint, where each term represents the cost associated with a specific type of ad:

• Television ads: \( \\)1000 \times x \)
• Radio ads: \( \\(200 \times y \)
• Newspaper ads: \( \\)500 \times z \)

Adhering to a budget constraint is essential for any business to ensure financial sustainability. In this case, the equation helps maintain financial discipline by limiting ad spending to exactly \( \$42,000 \), requiring a careful distribution of funds amongst the channels to optimize ad exposure while respecting the budget.

Variable Substitution

Variable substitution is a technique used to simplify systems of equations, making them easier to solve. By substituting one variable's expression from one equation into another, we reduce the number of variables involved, simplifying the system.
For instance, in our problem, the equation \( x = y + z \) allows us to write \( x \) in terms of \( y \) and \( z \). Substituting \( x = y + z \) into other equations reduces the complexity of the system:

• The first equation becomes \( 2y + 2z = 60 \), simplifying to \( y + z = 30 \).
• The budget constraint becomes \( 1000(y + z) + 200y + 500z = 42000 \), simplifying to \( y + 2z = 42 \).

These steps transform the original system into one with just two equations involving two variables, \( y \) and \( z \). This reduction process is crucial because solving smaller systems requires less computational effort and is generally more straightforward, aiding in efficiently reaching a solution.