How many of each type of ad can the department run each month?

Question

A health insurance company advertises on television, on radio, and in the local newspaper. The marketing department has an advertising budget of $87,000 per month. A television ad costs $2000, a radio ad costs $200, and a newspaper ad costs $600. The department wants to run 70 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?

Answer 1

This is a system of equations problem that consists of 3 variables. We can choose what variables represent the different forms of advertisements, so let's choose something easy to represent each.

Let T = number of television ads, R = number of radio ads, and N = number of newspaper ads

The second sentence tells us that the total cost of the advertisements is $87,000 per month, and the third sentence tells us how the price to run each type of advertisement. With this in mind, we can set up an equation that relates the price to the number of each advertisement type.

1) 2,000T + 200R + 600N = 87,000

The following sentence states that the department wants to run 70 ads per month, and so if we add up the numbers of each advertisement type, we can get our second equation:

2) T + R + N = 70

Lastly in the same sentence, it tells us the number of television sets is equal to the number of radio and newspaper ads combined (add them).

3) T = R + N

With these three equations in mind, we can use substitution or elimination to solve for the values of each variable, I will use substitution in this case.

We can start by substituting the T from Equation 3 into the T for Equation 2 and solve for R:

(R+N) + R + N = 70

2R + 2N = 70

2R = 70 - 2N

4) R = 35 - N

Taking a step back for a moment, if we look at Equation 1, we can substitute the T with Equation 2 in a similar fashion:

2,000(R + N) + 200R + 600N = 87,000

2,000R + 2,000N + 200R + 600N = 87,000

5) 2,200R + 2,600N = 87,000

Now we simply combine what we found with Equation 4 and Equation 5 by substituting the R which gives us the value for N:

2,200(35 - N) + 2b600N = 87,000

77,000 - 2,200N + 2,600N = 87,000

400N = 10,000

N = 25

Now that we know N, we can solve for R back in Equation 4:

R = 35 - 25

R = 10

Lastly, we can simply solve Equation 3 using the values of R and N we found:

T = 10 + 25

T = 35

We can double-check if our values agree with the expected numbers in Equation 1 and Equation 2, to verify our answer is correct. I hope this helps!