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A hotel finds that its total annual revenue and the number of rooms occupied by guests can be best modeled by the function R = 3 log (n^2 + 10n) , n > 0 where R is the total annual revenue in, millions of dollars, and n is the number of rooms occupied daily by guests. The hotel needs an annual revenue of 12 million to be profitable. Calculate the minimum number of rooms that must be occupied daily for the hotel to be profitable.

User Strille
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1 Answer

4 votes

Answer:

95 rooms

Explanation:

R = 3log(n^2 + 10n)

R = 12

3log(n^2 + 10n) = 12

Divide both sides by 3

Log(n^2 + 10n) = 4

Take log off both sides

n^2 + 10n = 10^4

n^2 + 10n - 10000 = 0

Solve the equation above using the quadratic formula

n = [-b + or - √(b^2 - 4ac)] ÷ 2a

The value of n must be positive

Therefore, n = [-b + √(b^2 - 4ac)] ÷ 2a

a = 1, b = 10, c = -10000

n = [-10 + √(10^2 - 4(1)(-10000)] ÷ 2(1) = (-10 + √40100) ÷ 2 = (-10 +200.25) ÷ 2 = 190.25 ÷ 2 = 95.125

Therefore, minimum number of rooms that must be occupied daily for the hotel to be profitable is 95

User Noamt
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