My Math Forum Changing rent rates

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 November 20th, 2015, 02:33 AM #1 Newbie   Joined: Nov 2015 From: Los Angeles Posts: 2 Thanks: 0 Changing rent rates I don't know if this is a calculus problem or what but here it is.... Imagine that I'm charging $3,250 for 7 days of rent and$6,100 for 14 days. Rent gets gradually cheaper the longer you stay. I want to figure out how much to charge for day 8, day 9, day 10, day 11, day 12 and day 13. It's not as simple as adding $464.28 more for day 8 because by the time you get to day 13 it's almost$6,500 -- That's MORE than the $6,100 cost of the 14 day rate. There's gotta be some sort of formula for this. Like multiplying$464.28 times .995 and then and then adding that number to $3,250 and then repeat for the next day.  November 20th, 2015, 06:38 AM #2 Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 This is very unclear. Do you mean that the total rent for days 1 through 7 is$3250 and the total rent is $6100 for days 1 through 14? And you want a single formula, say, r(d), where "d" is the day from 1 to 14 and "r(d)" is the rent for day "d"? So you want to have$\displaystyle \sum_{d= 1}^{7} r(d)= 3250$and$\displaystyle \sum_{d=1}^{14} r(d)= 6100$. There are, in fact, many different formulas that would work. Since we have two conditions to satisfy, the simplest such formula would be "r(d)= ad+ b" for sum numbers, a and b. The first condition becomes [a+ b]+ [2a+ b]+ [3a+ b]+ [4a+ b]+ [5a+ b]+ [6a+ b]+ [7a+ b]= 28a+ 7b= 3250. The second condition would be the same with the coefficients of a going from 1 to 7. We can simplify it a bit by using the fact that the first 7 terms are the same as above so we have 3250+ [8a+ b]+ [9a+ b]+ [10a+ b]+ [11a+ b]+ [12a+ b]+ [13a+ b]+ [14a+ b]= 3250+ 77a+ 7b= 6100 which is the same as 77a+ 7b= 6100- 3250= 2850. So we need to solve the two equation 77a+ 7b= 2850 and 28a+ 7b= 3250. Since both equations involve "7b" we can eliminate b by subtracting the second equation from the first: 77a- 28a= 49a= 2850- 3250= -400. Dividing both sides by 49, a= -400/49= -8.16 (to two decimal places). Putting that into 28a+ 7b= 3250, 28(-8.16)+ 7b= 228.5+ 7b= 3250 so 7b= 3250- 228.5= 3021. One possible formula is r(d)= 3021- 8.16d.  November 20th, 2015, 12:13 PM #3 Newbie Joined: Nov 2015 From: Los Angeles Posts: 2 Thanks: 0 Total rent for 7 days is$3,250. Total rent for 14 days is \$6,100 For some reason my text keeps getting scuttled by this site. I'll have a look at your equation. I've actually set up an OpenOffice spreadsheet whereby I can quickly insert a number by which the price of rent declines for each day by the fixed multiple (number) that I enter. I found that 0.99 is the closest number that works. I'm wondering if there's a formula that would determine that multiplier if all we know is the total cost of rent for 7 days and the total cost for 14 days. Of course I could always do it the trial and error way (by testing entering numbers within OpenOffice). I'm just wondering if there's a formula that would give me the exact number. Last edited by HeadScratcher; November 20th, 2015 at 12:27 PM.
 November 20th, 2015, 02:35 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 A little latex tip: You need dollar signs at the beginning AND the end of your code. Right now, it's rendering the text between the two dollar signs as latex.

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