Question

A company has $13,350 available per month for advertising. Newspaper ads cost $240 each and can't run more than 23 times per month. Radio ads cost $500 each and can't run more than 35 times per month at this price. Each newspaper ad reaches 6100 potential customers, and each radio ad reaches 7200 potential customers. The company wants to maximize the number of ad exposures to potential customers. Use n for number of Newspaper advertisements and r for number of Radio advertisements

Answer 1

1. The optimization model for the total ad exposures is Maximize: 6100n + 7200r.

2. The constraints are subject to the following:

Newspaper ads: 40n ≤ 13350 n ≤ 23

Radio ads: 500r ≤ 13350 r ≤ 35.

An optimization model is a mathematical technique that can find the best solution to a problem from a set of possible choices, taking into account some specific constraints and objectives.

The total monthly available funds for advertising = $13,350

The cost of newspaper ads per unit = $240

The total number of newspaper ads required per month, n ≤ 23

The cost of radio ads per unit = $500

The total number of radio ads required per month, r ≤ 35

The number of potential customers reached by each newspaper ad = 6,100

The number of potential customers reached by each radio ad = 7,200

To solve this problem, we can set up the following optimization model:

Maximize: 6100n + 7200r (total ad exposures)

Subject to:

Cost constraint for newspaper ads: 240n ≤ 13350 n ≤ 23

Cost constraint for radio ads: 500r ≤ 13350 r ≤ 35

Thus, we maximize the total ad exposures while staying within the cost and frequency constraints.