Question

A hardware store will run an advertising campaign using radio and newspaper. Every dollar spent on radio advertising will reach 60 people in the "Under $35,000" market, and 60 people in the "Over $35,000" market. Every dollar spent on newspaper advertising will reach 100 people in the "Under $35,000" market, and 20 people in the "Over $35,000" market. If the store wants to reach at least 210,000 people in the "Under $35,000" market and 240,000 people in the "Over $35,000" market, how much should it spend on each type of advertising to minimize the cost?

Minimum amount spent on advertising (in dollars):

Dollars spent on radio advertising

Dollars spent on newspaper advertising

Answer 1

We will solve over problem by simplex method as follows :

Let x1= dollar spent on radio Advertisment under $35000

x2= dollar spent on the radio advertisment over $35000

x3= dollar spent on the newspaper Advtisment under $35000

x4=dollar spent on the newspaper Advtisment over $35000

Then problem is

Min Z= x1+x2+x3+x4

Subjected to constraints

60x1+100x3 >=210000

60x2+20x4>=240000

x1,x2,x3,x4>=0

Solution is atteched below

thus total minimum cost z=$6100

Dollar spent on radio Advtisment=x1+x2=$4000

Dollar spent on newspaper Advtisment=x3+x4=$2100