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# I have posted several question i need help SAP

wrote...
Express Lane Plus
Posts: 485
Rep: 2 0
3 weeks ago
 I have posted several question i need help SAP 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 Attached file(s) Thumbnail(s): You must login or register to gain access to these attachments. Read 65 times 3 Replies
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wrote...
Educator
Top Poster
Posts: 24976
3 weeks ago Edited: 3 weeks ago, bio_man
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wrote...
Staff Member
3 weeks ago
 • Solution:Let R= the revenue function = quantity × priceLet n= the number of times the price is reduced by $7 dollars. Then:(price =$9 − n · ($7) = 9 − 7n dollars,quantity = number of car wash = 29000 car wash + n · (32000 car wash) = 29000+ 32000n car washHence R(n) = (29000+ 32000n)(9 − 7n) = 261000 +85000n -224000n2 to maximize (for 0 ≤ n ≤ 40). R'(n) = 85000− 448000n = 0 ⇒ n =85000/448000 = 0.19 is the only critical number R''(n) = −448000 ⇒ R''(0.19) = −448000 < 0The best fare to maximize the revenue is then:$ 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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