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wrote...
Express Lane Plus
Posts: 485
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3 weeks 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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wrote...
Educator
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Posts: 24976
3 weeks ago Edited: 3 weeks ago, bio_man
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I replied to most of them already, except for this one about revenue. You need to create a revenue function, find its derivative, set it equal to zero, find the critical points and sub them back into the original function, this will give you the maximum revenue... currently away from computer

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wrote...
Staff Member
3 weeks 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
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wrote...
Educator
3 weeks 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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