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spayn [35]
3 years ago
14

The Sea Wharf Restaurant would like to determine the best way to allocate a monthly advertising budget of $1000 between newspape

r advertising and radio advertising. Management decided that at least 25% of the budget must be spent on each type of media, and that the amount of money spent on local newspaper advertising must be at least twice the amount spent on radio advertising. A marketing consultant developed an index that measures audience exposure per dollar of advertising on a scale from 0 to 100, with higher values implying greater audience exposure. If the value of the index for local newspaper advertising is 50 and the value of the index for spot radio advertising is 80, how should the restaurant allocate its advertising budget in order to maximize the value of total audience exposure?
a. Formulate a linear programming model that can be used to determine how the restaurant should allocate its advertising budget in order to maximize the value of total audience exposure.
b. Solve the problem using the graphical solution procedure.
Mathematics
1 answer:
Sati [7]3 years ago
4 0

Answer:

Step-by-step explanation:

to find the answer you will want to gather the indenpent varible and dependent varrible. SO when you look at the problem you see 25%. You have to find the amount of money on each thing they bought. The budget and how much  they have to spend.

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<h2>Answer:</h2>

The following which is not a requirement of a standard form of equation Ax+By=C is:

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<h2>Step-by-step explanation:</h2>

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3 years ago
6/(x+1)-5/2=6/(3x+3)
jolli1 [7]
If you would like to calculate 6/(x+1)-5/2=6/(3x+3), you can do this using the following steps:

6/(x+1)-5/2=6/(3x+3)
6/(x+1)-5/2=6/(3(x+1))      /*(x+1)
6 - 5/2 * (x+1) = 6/3
6 - 2 = 5/2 * (x+1)
4 = 5/2 * (x+1)     /*2/5
4 * 2/5 = x + 1
8/5 - 1 = x
x = 8/5 - 5/5 = 3/5

The correct result would be 3/5.
6 0
3 years ago
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