The point of this is to create a linear programming model p

The point of this is to create a linear programming model ( pretty much put the information given in equatios) but not solve for them

The manager of a department store in seattle is attempting to decide on the types and amounts of advertising the store should use. He has invited representatives from the local radio station, television station, and newspaper to make presentations in which they describe their audiences. The television station representative indicates that a tv commercial, which costs $15,000, would reach 25,000 potential customes. The breakdown of the audience is as follows

male senior: 5000 female senior 5000

young male: 5000 young female 10000

the newspaper representative claims to be able to provide an audience of 10000 potential customers at a cost of $4000 per ad. The breakdown of the audience is

senior male 4000 . senior female 3000

young male 2000  young female 1000

the radio station representative says the audience for one of the stations commercials which cost $6000 is 15,000 customers . The breakdown is as follows

Senior male 1500 Senior female 15000

young male 4500 young female 7500

The store has the following advertising policy

use at least twice as many radio commercials as news paper ads

reach at least 100,000 customers

reach at least twice as many young people as senior citizens

make sure that at least 30% of the audience is female

Solution

Let the number of tv ads, newspaper ads and radio ads be:

t, x, & y respectively.

As advertising expenditure is a cost, we shall be required to minimize it. Therefore, the problem statement of the linear programming problem shall become:

Min. C = 15000*(t) + 4000*(x) + 6000*(y) $

S.T. y >= 2*(x)

or, -2*(x) + y >= 0 [This is the first constraint, viz. use at least twice as many radio commercials as newspaper advertisements];

25000*(t) + 10000*(x) + 15000*(y) >= 100000 [The second constraint, about reaching 100000 customers];

or, 5*(t) + 2*(x) + 3*(y) >= 20;

Then, the third constraint:

15000*(t) + 3000*(x) + 12000*(y) >= 2*{10000*(t) + 7000*(x) + 3000*(y)}

or, -5000*(t) -11000*(x) + 6000*(y) >= 0

or, -5*(t) -11*(x) + 6*(y)>=0

Finally, the fourth constraint is:

15000*(t) + 4000*(x) + 9000*(y) >= 0.3*{25000*(t) + 10000*(x) + 15000*(y)

or, 15000*(t) + 4000*(x) + 9000*(y) >= 7500*(t) + 3000*(x) + 4500*(y);

or, 7500*(t) + 1000*(x) + 4500*(y) >= 0.

or, 15*(t) + 2*(x) + 9*(y) >= 0.

Also, t >= 0;

x >= 0;

& y >= 0.