Could someone please help me figure out the equation for this math word problem?

In most businesses, increasing prices of their product can have a negative effect on the number of customers of the business. A bus company in a small town has an average number of riders of 1,000 per day. The bus company charges $2.00 for a ride. They conducted a survey of their customers and found that they will lose approximately 50 customers per day for each $.25 increase in fare.





This is all that I was given. Thanks so much to whomever can help.

Could someone please help me figure out the equation for this math word problem?
Current revenue is $2.00 * 1,000.





For every 25 cent increase, there will be a 50 person decrease in ridership.





So, in that case revenue would be ($2.00 + $0.25x) * (1,000 - 50x).





So, I would use R = (2+0.25x)(1000-50x) = -12.5x² +150x + 2000





Are you in calc or pre-calc? Do you know if you need to determine what the maximum fare increase is before revenue starts going down? (That's a typical type of question that's usually asked with these types of questions.)





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Reply:If x = fare price, and y = riders per day





y = -200x+1400





I found this out by take the first point (2,1000)





If the fare increasing causes a subtraction of riders, then it's a negative slope.





To get the y-intercept, just play it through til you figure out how many customers if the price was free (x=0). If we know 50 customers is lost for $0.25, then 200 customers are lost for every $1. Therefore, 400 customers are lost for an increase of $2. Equally the same by stating 400 customers are gained by decreasing the price by $2.





So with the values of (0,1400) and (2,1000). Simply get an equation from that.





The y-intercept is going to be 1400. The slope is the change in y divided by the chance in x. So that's a change of -400 for 2 x's, or -200 per x. So the slope is -200. Forming the final equation of:





y = -200x+1400
Reply:This price per fare can be raised to $3.50 , based on the information given before the profits begin to decrease again. At $3.50 the company will have 700 passengers a day. $3.50 times 700 = $2450.00. That is an increase of $450 per day for the company.
Reply:OK, whats the question?





So far all you've got is a series of statments and a relationship.(for every $.25 increase the Average Riders [we'll call it A(r)] decreases by 50.





Whats interesting in this problem is that revenue increase at each step (at a price of $3.00 you've lost an average of 200 riders per day, so you are not serving your community, but you are making more money $2400 Vs $2000), so the returns are not diminshing they are increasing. Once you get to 500 ppl and $6.00 the returns start to diminsh again, from a high point of $3000 per month - or was that the question?





It's anywhere from 600ppl at $5.00 to 500 ppl at $6.00 both are $3000 PM revenue, so that's your "Tipping Point".





But, we still need a goal for the question, unless I'm being dense
Reply:New number of customers = 1000 per day -50 x new price increase/$.25





C = 1000 - 50 (Price - 2.00)/.25





example : If the bus fare is raised to $2.50 then





C = 1000 - 50($2.50 - $2.00 ) / .25


C= 1000 -100


C= 900 users per day
Reply:I think they are ok until they charge $5.25. It is the first time that their rev. falls under the original revenue of $2,000. Which is where they started charging $2.00 to 1,000 riders. Can you use an excel spreadsheet? That is how I did it. Otherwise just get out a calculator ,pencil and paper and start listing. price x # of riders. Greed is good!! For a while any way.





price riders revenue


$2.00 1000 $2,000.00


$2.25 950 $2,137.50


$2.50 900 $2,250.00


$2.75 850 $ 2,337.50


$3.00 800 $2,400.00


$3.25 750 $2,437.50


$3.50 700 $2,450.00


$3.75 650 $2,437.50


$4.00 600 $2,400.00


$4.25 550 $2,337.50


$4.50 500 $2,250.00


$4.75 450 $2,137.50


$5.00 400 $2,000.00


$5.25 350 $1,837.50


$5.50 300 $1,650.00


$5.75 250 $1,437.50


$6.00 200 $1,200.00
Reply:well, you still end up earning more $ with the rate increase....even with the 50 lost customers per $.25 rate increase. So it's bad you lose customers with the hike in fares but good that you profit more even after losing those customers.
Reply:What is your question?








No. of Customer $1,000X $ 2 =2,000


lose 50 per day X $2fare = 100
Reply:Let the number of increases be x





The new price for a ride is 2.00 + 0.25x





The new number of sales is 1000 - 50x





The total revenue is (2 + 0.25x)(1000 - 50x) = -12.5x^2 + 150x +2000