Number of Newspaper ads to run is? Number of Radio ads to run is? Maximum target group exposure is?

Question

A company has $9,510 available per month for advertising. Newspaper ads cost $100 each and can't run more than 23 times per month. Radio ads cost $520 each and can't run more than 31 times per month at this price.

Each newspaper ad reaches 5100 potential customers, and each radio ad reaches 5750 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 .

Answer 1

Hello! You can use Wolfram Alpha's linear optimization calculator to solve this problem. Here is the link:

https://www.wolframalpha.com/widgets/view.jsp?id=daa12bbf5e4daec7b363737d6d496120

Once you reach this link, you will see sample constraints. Where it says "optimize," go to the drop down menu and select "Max", because this is a company trying to maximize the number of potential customers that will see the product.

Let x_1=n, the number of newspaper ads, and x_2=r, the number of radio ads. Where it says objective function, delete the sample function and put in 5100x_1+5750x_2 (no comma needed at the end). This sum represents the total number of potential customers that will see your product based on the number of newspaper ads you choose to buy and the number of radio ads you choose to buy. You want this sum to be as high as possible given the constraints.

In the area where it says "subject to," you will see three sample functions:

0.3x_1 + 0.1x_2 <= 2.7,

0.6x_1 + 0.4x_2 >= 6, (delete this one)

0.5x_1 + 0.5x_2 = 6, (also go ahead and delete this one)

Where it says 0.3x_1 + 0.1x_2 <= 2.7, delete everything in that box, and replace it with 100x_1+520x_2 <= 9510, (put the comma in there too!) This function represents how much money will be spent as a function of how many newspaper ads and how many radio ads you buy.

Where it says "and: x_1>=0, x_2>=0" delete everything in that box and replace it with the constraints "0<=x_1<=23, 0<=x_2<=31" (copy and paste what you see inside the quotation marks exactly as you see it, except for the quotation marks themselves.)

After you've put everything in, hit the blue "solve" button. The calculator's answer will be (23, 721/52). In other words, the optimal number of newspaper ads to buy is 23, and the optimal number of radio ads to buy is 721/52 (in decimal form that's 721/52=13.865). However, you can only buy a number of ads represented by a whole number, so 13.865 is not possible. Also, if you round up to 14, you will exceed your cost constraint. So you must round down.

Your final answer will be 23 newspaper ads and 13 radio ads as the optimal combination. I hope that helped!