Chapter 8.8, Problem 23E

To determine

To find:

How many newspaper ads and radio ads you should run each month?

Answer to Problem 23E

You should run 30 newspaper ads and 20 radio ads each month.

Explanation of Solution

Given:

You own a business that advertises in a local newspaper and over the radio. A newspaper ad costs $600. A radio ad costs $300. You have a monthly budget of $24,000 and want to run 50 ads each month.

Calculation:

Let x represent number of newspaper ads and y represent number of radio ads.

We have been given that you want to run 50 ads each month, so number of radio ads and newspaper ads should be equal to 50. We can represent this information in an equation as:

  $x+y=\mathrm{50...}(1)$

The cost of each newspaper ad is $600, so cost of xnewspaper ads would be $600x$ .

The cost of each radio ad is $300, so cost of y radio ads would be $300y$ .

Since the monthly budget is $24,000, so cost of newspaper ads and radio ads should be equal to $24,000.We can represent this information in an equation as:

  $600x+300y=24,\mathrm{000...}(2)$

Now we will use substitution method to solve our system of equations. From equation (1), we will get:

  $x=50-y$

Upon substituting this value in equation (2), we will get:

  $\begin{array}{l}600(50-y)+300y=24,000\\ 30,000-600y+300y=24,000\left(\text{Using distributive property}\right)\\ 30,000-300y=24,000\left(\text{Combining like terms}\right)\\ 30,000-30,000-300y=24,000-30,000\left(\text{Subtacting }30,000\text{from both sides}\right)\\ -300y=-6,000\\ \frac{-300y}{-300}=\frac{-6,000}{-300}\left(\text{Dividing both sides by }-300\right)\\ y=20\end{array}$

Therefore, you should run 20 radio ads each month.

Now we will substitute $y=20$ in equation (1) as:

  $\begin{array}{l}x+20=50\\ x+20-20=50-20\\ x=30\end{array}$

Therefore, you should run 30 newspaper ads each month.