Two advertising media are being considered for promotion of a product. Radio ads cost $400 each, while newspaper ads cost $600 each. The total budget is $7,200 per week.
The total number of ads should be at least 15 and at least 2 ads of each type (Radio and
newspaper) should be considered. Each newspaper ad reaches 6,000 people, while each
radio ad reaches 2,000 people. The company wishes to reach as many people as possible
while meeting all the constraints stated.
1.
Write the constraints and the objective function for this problem?
(3 pts)
2.
How many ads of each type should be placed?
(5 pts)

1. 1. The constraints are as follows -

a) Limited weekly budgets

b) Limited ads requiring that there should be at least 15 and at least 2 ads of each type (Radio and  newspaper)

c) Newspaper ad being costly reach larger population segment as compared to the radio ads which is cheaper than newspaper ad but reaches only a small portion of population

2. 6 newspaper adds and 9 radio adds

Step-by-step explanation:

1. The constraints are as follows -

a) Limited weekly budgets

b) Limited ads requiring that there should be at least 15 and at least 2 ads of each type (Radio and  newspaper)

c) Newspaper ad being costly reach larger population segment as compared to the radio ads which is cheaper than newspaper ad but reaches only a small portion of population

The objective function

Maximum number of people per week = 6000  * number of newspapaer ads + 2000 * number of radio ads

b) $7,200

Minimum radio adds = 3 this will cost 3 * $400 = $1200

Remaining money = $7200 - $1200 = $6000

Cost of one add of news paper  = $ 6000/$600 = 10

But the minimum ads need to be 15

So, If Newspaper adds are reduced to 6, the radio adds will be 9 giving a sum of 15

2nd one in the first row is correct