Toll Road Profit
A toll road in Puzzletown averages 300,000 cars a day when the toll is $2.00 per car. A detailed study has proven that for every $.10 increase in the toll, 10,000 fewer cars will use the road each day.
What toll maximized the revenue?
What toll maximized the revenue?
posted by Zaux at
6:54 AM
13 Comments:
$500,000. Then number of cars is then 250,000 and the toll is $2.50.
$2.50 Toll
250000 Avg Cars
$625,000 revenue
Hi Chris ...
The $2.50 toll is the correct amount to maximize profits.
I've given this some more thought....
To maximise profits, maybe te toll is $3.00, as you will only have 200,000 cars going through, and the same revenue as you have at the $2.00 toll, requiring less staffing costs...?
Another senior moment for me. No idea why I typed $500,000. I agree with Karl's value.
Here's how I did it:
T = 2 - 0.1*(N-300000)/10000
R = NT, where T is tol, N is #cars, R is revenue
T = 2 - N/100000 + 0.1*300000/10000
T = 5 - N/100000
R = N(5 - N/100000) = 5N - N²/100000
This will be a maximum when the rate of change of R with respect to N is 0.
dR/dN = 5 - 2N/10000 = 0 (at max R)
So 5 = 2N/100000 => N = 250000.
=> T = 5 - 250000/100000 = 2.5
and R = 2.5*250000 = 625000.
The other way to see the maximum, is to observe that quadratics are symetrical about their centre.
R = N(5 - N/100000).
R = 0 when N = 0 and when N = 500000.
So the midpoint is N = 250000, as before.
Solution:
n = number of cars in units of 1000
t = the toll, in dollars
x represents number of dimes to be charged in excess of $2.00.
Toll will be:
t = 2.00 + 0.10x
Since the number of cars will decrease by 10,000 for each unit increment in x, the daily number of cars using the road will be:
R(x) = n*t
= (300-10x)(2.00+0.10x)
= 600+10x-x^2
R!(x)= 10-2x
0 = 10-2x
x = 5
The toll should be increased by 50 cents (5 x 10 cents) to $2.50
I followed this published solution okay until I reached the point of factoring:
600+10x+x^2
So, thanks to Chris previously mentioning WolframAlpha, I ran it through there.
WolframAlpha got -5.
I hoping one of you math wizards will explain why the published solution was "5" and WolframAlpha got "-5".
Sorry ... previous post is mine (Zaux)
Hi Zaux. Your stated definition of x is peculiar. Your x is (30000-n)/n.
Wolframalpha gives x = 20 and +30.
You had already had 600 + 10x - x^2 in factor form (300 - 10x)(2.00 + 0.10x) and that also gives x = 30 and -20. The midpoint is x = 5.
You also chnged a sign in one version of the quadratic (possibly a simple typo).
I've no idea where your -5 came from.
Oh ... OK ... thanks chris
ooops. I meant wolframalpha gave -20 and +30 for x.
...aaaarg, I also meant that your x is (300000-n)/10000.
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