Question

A candidate for mayor in a small town has allocated $40,000 for last-minute advertising in the days pre- ceding the election. Two types of ads will be used: radio and television. Each radio ad costs $200 and reaches an estimated 3,000 people. Each television ad costs $500 and reaches an estimated 7,000 peo- ple. In planning the advertising campaign, the cam- paign manager would like to reach as many people as possible, but she has stipulated that at least 10 ads of each type must be used. Also, the number of radio ads must be at least as great as the number of televi- sion ads. How many ads of each type should be used? How many people will this reach?

Answer 1

Answer : To reach the maximum number of people by adds the candidate of mayor should spent $40,000 equally on both radio add and television add. This means that the candidate of mayor should spent $20,000 on radio add and $20,000 on television add.

Now,

$200 is the cost for 1 radio add

=> $1 is the cost for = 1 / 200 radio add

=> $20,000 is the cots for = (1 * 20000) / 200 = 100 radio add.

Therefore, 100 radio adds are possible by spending $20,000.

As 1 radio add reaches to 3000 people, so, 100 radio adds will reach to (100 * 3000) = 300,000 people.

Again,

$500 is the cost for 1 television add

=> $1 is the cost for = 1 / 500 television add

=> $20,000 is the cots for = (1 * 20000) / 500 = 40 television add.

Therefore, 40 television adds are possible by spending $20,000.

As 1 television add reaches to 7000 people, so, 40 television adds will reach to (40 * 7000) = 280,000 people.

Therefore, here the candidate of mayor should use 100 radio adds and 40 television adds.

100 radio adds will reach to 300000 people and 40 television adds will reach to 280,000 people.