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 cx 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

Combined cost C(x) = cx + 4c/(x+1)

minimum C(x) ==> C '(x) = 0

==> c - 4c/(x+1)2 = 0

==> 4c/(x+1)2 = c

==> (x+1)2 = 4

==> x +1 = 2

==> x = 1 mile

Hence she should live 1 mile from work to minimise comined costs.

Step-by-step explanation:

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total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98