Help with a quick grade 10 word problem?
The Sudbury Transit Sytem carries 15,000 bus riders per day for a fare of $1.90. The city wishes to increase ridership in an effort to reduce car pollution, while at the same time maximizing the Transit revenues. The city conducted a survey and the results indicated that the number of riders would increase by 500 for every $0.05 decrease in fare.
a) Write an equation to represent the daily revenue.
b)What fare will produce the greates revenue? What is the expected daily ridership?
c)The city calculates that it must take in $24,675 per day to break even. What ticket prices would enable the city to earn a profit?
****Please show all work in steps so I can follow through!!****
Comments
a. "y" is revenue, "x" is the number of times they decrease the price by $0.05
y = (1.90 - 0.05x)(15000 + 500x)
y = -25x² + 200x + 28500
b. By graphing the equation, you can find the maximum point is (4, 28900), which means that you should multiply 4*.05 and subtract it from 1.90.
1.90 - (4*0.05)
1.90 - 0.20
$1.70 is the answer
c. Set 24675 equal to the equation and solve.
24675 = -25x² + 200x + 28500
0 = -25x² + 200x + 3825
0 = -25(x - 17)(x + 9)
x - 17 = 0 and x + 9 = 0
x = 17 and x = -9
Now, sub these two values into the 1.90 - 0.05x equation to get the numbers:
1.90 - (17*0.05)
1.90 - 0.85
$1.05
1.90 - (-9*0.05)
1.90 + .045
$2.35
Any price between (not equal to) $1.05 and $2.35 will give the city a profit.
by the way, im in 8th grade and this was really easy
a)let the number of times that u reduce the fare by .05 as x. Your revenue is the number of riders times the fare. the revenue can be expressed as (15000+500x)(1.9-.05x).
b)u can use FOIL to write the function as -25x squared +200x+28500. This means ur a b and c values are respectively -25, 200, and 28500. If you graphed the quadratic function, you would get a parabola facing down. the point where the revenue is the highest is the maximum. The maximum is the y value of the parabola when the x is the axis of symmetry. The axis of symmetry is represented by -b/2a, which is
-200/-50 or 4. the fare is 1.9-.05x or $1.70. The number of riders is 500x +15000 or 17000.
c) your values of x between the 2 values that give 24675 will have greater revenue than 24675. The two values of x the give 24675 are found by the quadratic formula. they are -30 and 38. This means all values of x below 38 and above -30 will work. This means all fares between $3.40 and $0.00 will work.
I agree with "Big"'s calculations.
I only add that to find the greatest revenue, I took the derivative of the function -25x² + 200x + 28500, which is -50x + 200 and set it to zero to obtain x=4. At this point the revenue function has a min or max. (The derivative of -50x + 200 is -50. That's negative, so it's a max.)
So the fare that produces the greatest revenue is $1.09 - 4*.05=$1.70.
The expected ridership is 15000+4*200 = 17000.
take a calendar. now take it off the wall. placed it on the table. decrease little squares of paper with the numbers one million-30. set up the squares over the calendar so as that there are 5 sundays and september 2 is a school day. in spite of day the thirtieth falls on is your answer.