Two advertising media are being considered for the promotion of a product. Radio ads cost $480 each, while newspaper ads cost $550 each. The total budget is $20,000 per week. The total number of ads should be at least 30, with a max of 5 newspaper ads. Each newspaper ad reaches 8,000 people, while each radio ad reaches 5,000 people. Let R = # of radio ads
Let P = # of newspaper ads
Max 5000 R + 8000P
s. t.
480R + 550P <= 20000 cost of ads
R + P >= 30 total # of ads
P <= 5 max number of newspaper ads
R, P >= 0
Round your answers to the highest whole numbers
The company wishes to reach as many people as possible while meeting all the constraints stated, what is maximum reach?
How many ads of each type should be placed?

Rounding the solution:

x₁  =  37

x₂  =  4

z(max)  =  217000

Step-by-step explanation:

                                                             Cost $

Promotion on Radio R           (x₁)           480

Promotion on Newspaper P (x₂)           550

Availability                                           20000

z  =  5000*x₁   +  8000*x₂      to maximize

Subject to

Budget          20000

480*x₁  +  550*x₂   ≤  20000

Maximum number of ads   30

x₁   +   x₂   ≥  30

Maximum number of newspaper ads  5

x₂  ≤  5

The Model:

z  =  5000*x₁   +  8000*x₂      to maximize

Subject to:

480*x₁  +  550*x₂   ≤  20000

x₁   +   x₂   ≥  30

x₂  ≤  5

General constraints:

x ≥   0  ;  x  ≥ 0    both integers

NOTE: The model was considered that  x₁    and  x₂  as integers. The AtomZmath did not find an integer optimal solution. But in problem statement was established that " Round your answers to the highest whole number". I decided to pose the non-integer solution

After 6 iterations optimal solution is:

x₁   =  37.15

x₂  = 3.925

z(max)  =  217150

Rounding the solution:

x₁  =  37

x₂  =  4

z(max)  =  217000

answer: there are 453.592 grams in a pound.

24+x=y

step-by-step explanation: