November 20th, 2015, 01:33 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, 05:38 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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, 11:13 AM  #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 11:27 AM. 
November 20th, 2015, 01:35 PM  #4 
Math Team Joined: Nov 2014 From: Australia Posts: 688 Thanks: 243 
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.


Tags 
changing, rates, rent 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Changing the intergrator  Robertoo  Calculus  1  June 19th, 2014 04:09 PM 
Changing the range of mod x  Ter  Algebra  1  April 19th, 2012 11:09 AM 
Rent problem!  DGillett  Elementary Math  3  June 18th, 2010 09:13 AM 
Related Rates Problem: Find rate that distance is changing.  DQ  Calculus  2  June 5th, 2010 04:34 AM 