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? Use the methods outlined in this section to find the minimum.

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

For first 30 GB, a fix payment of 500 Pesos, equation would be:

C(g) = 500(g⁰), note: g⁰ = 1, where g ≤ 30

It is also equal to C(g) = 500 when g ≤ 30.

For 30 GB to 49 GB, 30 pesos each GB will be charged, equation would be:  

C(g) = 500 + 30(g-30)  

C(g) = 500 + 30g - 90

C(g) = 30g + 410, where 30 < g < 50

For 50 GB and above, the equation will be:  

C(g) = 1000(g⁰), where g ≥ 50

It is also equal to C(g) = 1000, when g ≥ 50.