Question

A candidate for mayor in a small town has allocated $40,000 for last-minute advertising in the days preceding the election. Two types of ads will be used: radio and television. Each radio ad costs $200 and reaches an estimated 3,000 people. Each television ad costs $500 and reaches an estimated 7,000 people. In planning the advertising campaign, the campaign manager would like to reach as many people as possible, but she has stipulated that at least 10 ads of each type must be used. Also, the number of radio ads must be at least as great as the number of television ads. How many ads of each type should be used? How many people will this reach?Let X1= the number of radio ads purchasedX2= the number of television ads purchasedMaximize3,000X1+7,000X2(maximize exposure)Subject to:200X1+500X2≤40,000(budget constraint)X1≥10(at least 10 radio ads purchased)X2≥10(at least 10 television ads purchased)X1≥X2(# of radio ads ≥ # of television ads)X1, X2≥0(non-negativity constraints)

Answer 1

Answer:

• 175 radio ads
• 10 television ads
• 595,000 people

Step-by-step explanation:

Radio ads reach 3000/200 = 15 people per dollar.

Television ads reach 7000/500 = 14 people per dollar.

Except for the constraints on the number of TV ads, the best value for the advertising dollar comes from radio ads.

So, we must satisfy the constraint that 10 TV ads are the minimum. Then the remaining 35,000 in advertising budget can be spent on 175 radio ads. The number of people reached by this advertising will be ...

  175·3000 +10·7000 = 595,000 . . . people

10 TV ads and 175 radio ads should be used. This campaign will reach 595,000 people.

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This graph is drawn so the feasible region is white. Areas outside the feasible region are shaded. (This approach can make identifying the feasible region easier.) The object is to get the objective function line as far from the origin as possible. The feasible region vertex (175, 10) does that.