A department store chain has up to $22,000 to spend on television advertising fo
ID: 1103106 • Letter: A
Question
A department store chain has up to $22,000 to spend on television advertising for a sale. All ads will be placed with one television station, where a 30-second ad costs $1,000 on daytime TV and is viewed by 14,000 potential customers, $2,000 on prime-time TV and is viewed by 24,000 potential customers, and $1,500 on late-night TV and is viewed by 18,000 potential customers. The television station will not accept a total of more than 20 ads in all three time periods How many ads should be placed in each time period in order to maximize the number of potential customers who will see the ads? How many potential customers will see the ads? (Ignore repeated viewings of the ad by the same potential customer.) Include an interpretation of any nonzero slack variables in the optimal solution Select the correct choice below and fill in any answer boxes present in your choice. O A. The maxim 0 B. There is no way to maximize the number of potential customers. um number of potential customers who see the ads is people wheExplanation / Answer
Answer is option B.
There is no way to maximize the number of potential customers.
However, No. of ads placed in each time period can be calculated:
a) In day time: 22000$/1000$=22
but as the station cannot accept a total or more than 20 ads.
No. of ads displayed in day time will be =20 ads
b)In Prime time: 22000$/2000$=11
No. of ads displayed in prime time = 11ads
c)In late-night time: 22000$/1500$= 14.67
No. of ads displayed in late-night will be=15 ads
Hence, there is no way to maximise the number of potential customer.
In an optimal situation, slack variable is a variable which transforms any inequal expression into equal expression.
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