Maximize. Use n for number of Newspaper advertisements and r for number of Radio advertisements .

Question

A company has $4,910 available per month for advertising. Newspaper ads cost $130 each and can't run more than 19 times per month. Radio ads cost $310 each and can't run more than 28 times per month at this price.

Each newspaper ad reaches 6550 potential customers, and each radio ad reaches 7550 potential customers. The company wants to maximize the number of ad exposures to potential customers.

Maximize P= 6550n+7550r   subject to

n   ≤19

r   ≤28

130n+310r   ≤$4,910

Enter the solution below. If needed, round ads to 1 decimal palace and group exposure to the nearest whole person.

Number of Newspaper ads to run is 19 (It has to be 19)

Number of Radio ads to run is...

Maximum target group exposure is...  people

Answer 1

desmos.com/calculator/diz9msus4m

Answer 2

For maximum P, minimize n and maximize r. Minimum n =1 Compute corresponding maximum r using dollar constraint.

130 n + 310 r ≤ 4910 Substitute n =1

130 + 310 r ≤ 4910 solve for r = (4910 - 130)/310 = 15.42 round down to 15

Number of radio ad to run is 15

Maximum target group exposure P = 6550 x 1 + 7550 x 15 = 6550 + 113250 = 119800

I hope this helps.

Sincerely,

Shailesh (Sky) Kadakia

Expert tutor with WYZANT