Answer:
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