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Home Q & A Board Science-Related Homework Help Mathematics Calculus
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Catracho Catracho
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Posts: 529
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4 years ago
Hi i need urgent help







A car wash reduced the price of a basic wash as a promotion and test of the market. With the price at $9 the average monthly sales has been 29000. When the price dropped to $7, the average monthly sales rose to 32000. Assume that monthly sales is linearly related to the price
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bio_manbio_man
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#1
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4 years ago Edited: 4 years ago, bio_man
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#2
Staff Member
4 years ago
• Solution:
Let R= the revenue function = quantity × price
Let n= the number of times the price is reduced by $ 7 dollars. Then:
(
price = $9 − n · ($7) = 9 − 7n dollars,
quantity = number of car wash = 29000 car wash + n · (32000 car wash) = 29000+ 32000n car wash

Hence R(n) = (29000+ 32000n)(9 − 7n) = 261000 +85000n -224000n2      to maximize (for 0 ≤ n ≤ 40).

R'(n) = 85000− 448000n = 0 ⇒ n =85000/448000 = 0.19 is the only critical number     

R''(n) = −448000 ⇒ R''(0.19) = −448000 < 0

The best fare to maximize the revenue is then: $ 9 − 7(0.19) = $7.67 with 29000+ 32000(0.19) car wash = 35080 car wash and a revenue of R(0.19) = $269,064
wrote...
#3
Educator
4 years ago
I can't verify bolbol's answer, but the numbers are consistent with the questions range between 7 and 9. Very interesting take
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