The Fashion Store has 12000 available each month for adverti

The Fashion Store has $12,000 available each month for advertising. Newspaper ads cost $400 apiece and no more than 30 can be run per month. Radio ads cost $200 each and no more than 40 can run per month. TV ads cost $1,600 a piece, with a maximum of 7 available each month. Approximately 2000 women will see each newspaper ad, 1600 will hear each radio ad, and 15,000 will see each TV ad.

How much of each type of advertising should be used if the store wants to maximize exposure?

How many newspaper ads should be placed?

How many radio ads should be used?

How many TV ads should be run?

What is the maximum exposure? women

Solution

let newspaper ads = x

and radio ads = y

tv ads = z

inequalitites would be

x <= 30

y <= 40

400x + 200y + 1600z <= 12000

z <= 7

objective function is

P = 2000x + 1600y + 15000 z

Solving by simplex method:

Tableau #1
x      y      x      s1     s2     s3     s4     p           
1      0      1      1      0      0      0      0      30   
0      1      0      0      1      0      0      0      40   
400    200    400    0      0      1      0      0      12000
0      0      0      0      0      0      1      0      7    
-2000 -1600 -15000 0      0      0      0      1      0    

Tableau #2
x       y       x       s1      s2      s3      s4      p             
0       -0.5    0       1       0       -0.0025 0       0       0     
0       1       0       0       1       0       0       0       40    
1       0.5     1       0       0       0.0025 0       0       30    
0       0       0       0       0       0       1       0       7     
13000   5900    0       0       0       37.5    0       1       450000

Optimal solution : Maximum exposure 450000

let newspaper ads = x =0

and radio ads = y =0

tv ads = z = 30