A company has $\$ 9,730$ avallable per month for advertising. Newspaper ads cost $\$ 140$ each and can't run more than 23 times per month. Radio ads cost $\$ 370$ 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 6250 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. Maximize $P=$ subject to $\leq 23$ $\leq 31$ $\leq \$ 9,730$ Enter the solution below. If needed, round ads to 1 decimal palace and group exposure to the nearest whole person. Number of Newspaper ads to run is Number of Radio ads to run is Maximum target group exposure is people Question Help: $⿴$ Message instructor $D$ Post to forum
Solution
Step 1 ：First, we find the maximum number of newspaper ads we can afford with the total budget, which is \(\lfloor 9730/140 \rfloor = 69\). However, we can only run 23 ads, so \(n = 23\).
Step 2 ：Then, we subtract the cost of these ads from the total budget to find how much we have left for radio ads: \(9730 - 23*140 = 6730\).
Step 3 ：Next, we find the maximum number of radio ads we can afford with the remaining budget, which is \(\lfloor 6730/370 \rfloor = 18\). However, we can only run 31 ads, so \(r = 18\).
Step 4 ：Finally, we calculate the total number of potential customers reached: \(5100*23 + 6250*18 = 274500\).
Step 5 ：\(\boxed{\text{So, the company should run 23 newspaper ads and 18 radio ads to maximize the number of potential customers reached, which is 274500 people.}}\)
