Question

A company has $9,670 available per month for advertising. Newspaper ads cost $140 each and can't run more than 20 times per month. Radio ads cost $430 each and can't run more than 27 times per month at this price. Each newspaper ad reaches 5900 potential customers, and each radio ad reaches 6900 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 . Maximize P = subject to ≤ 20 ≤ 27 ≤ $ 9 , 670 Enter the solution below. If needed, round ads to 1 decimal palace and group exposure to the nearest whole person. Excel tip: The number of advertisements does not need to be a whole number.

Number of Newspaper ads to run is _

Number of Radio ads to run is_

Maximum target group exposure is_ people

Answer 1

Let the number of Newspaper ads and Radio ads per month be x and y respectively. Then 0 ≤ x ≤ 20 and 0 ≤ y ≤ 27.

The reach of the advertisements is P(x,y) = 5900x+ 6900y.

The cost of the advertisements is C(x,y) = 140x+430y. Hence 140x+430y ≤ 9670 or, 14x+43y ≤ 967.

A graph of the lines x = 0, x = 20, y = 0 , y = 27 and 14x+43y = 967 is attached. The feasible region is bound by the lines OA (x = 0), AB (14x+43y = 967), BC ( x = 20) and OC ( y = 0).

The point A is ( 0,22.488) , say (0,22.5) and the point C is (20,15.977) , say(20,16).

At A, P(x, y) = 5900*0+ 6900*22.5 = 155250 and at C, P(x, y) = 5900*20+ 6900*16= 118000+110400 = 228400.

Thus,

Number of Newspaper ads to run is 20

Number of Radio ads to run is 16

Maximum target group exposure is 228400 people

27 to,22.5) C20, 16) 나4+ 4주성 = 967 20 0 40 To