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Nikitich [7]
3 years ago
14

A woman wishes to rent a house within 9 miles of her work. If she lives x miles from her work, her transportation cost will be c

x dollars per year, while her rent will be 4c/( x + 1) dollars per year, where c is a constant taking various situational factors into account. How far should she live from work to minimize her combined expenses for rent and transportation? Use the methods outlined in this section to find the minimum.
Mathematics
1 answer:
I am Lyosha [343]3 years ago
5 0

Answer:

the woman has to live 1 mile from work to minimize the expenses

Step-by-step explanation:

Given the data in the question;

the distance within 9 miles ⇒ 0 < x > 9

Total costs Q = cx + 4c/( x + 1)

costs should be minimum  ⇒ dQ/dx = 0

⇒ d/dx [ cx + 4c/( x + 1) ] = 0

⇒ ( x + 1)² = 4

take square root of both side

√[ ( x + 1)² ] = √4

x + 1 = 2

x = 2 - 1

x = 1

Therefore, the woman has to live 1 mile from work to minimize the expenses

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Previous concepts

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Solution to the problem

Let X the random variable that represent the number of children per fammili of a population, and for this case we know the following info:

Where \mu=2.5 and \sigma=1.7

We select a sample of n =64 >30 and we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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\sigma_{\bar X} = \frac{1.7}{\sqrt{64}}= 0.2125

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