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An unlimited supply of gasoline is available at one edge of a desert 800 miles wide, but there is no source in the desert itself. A truck can carry enough gasoline to go 500 miles (this is called the "load"), and it can build its own refueling stations at any spot along the way.
What is the minimum amount of gasoline (in loads) the truck will require in order to cross the desert?
Last edited by uzurpatorul (2009-06-27 06:23:59)
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with n = # of loads of fuel
when n = 3 then 766.666 miles are traveled.
when n = 4 then 838.095 miles are traveled.
So the amount of fuel required is between 3 and 4 loads.
We need an analytical form for the series but it is complicated and involves the psi function. We make do with an approximation to the sum.
A)
Solving for n:
n = 3.4435 loads of fuel necessary. When I bring this answer to the site, it still doesn't work. Possibly the approximation of the sum used in A is causing the error. When I sum the series exactly I get n = 3.4315, pretty close. This is out of Numerical Analysis by Francis Scheid.
Last edited by bobbym (2009-06-25 22:13:20)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Someone who has the ability may want to change the title from Crossing the dessert to Crossing the desert.
I'll be here at least once every decade.
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You general solution is right, however final answer is little over .
P.S.
I guess I was craving for sweets when I spelled the title.
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Hi uzurpatorul;
Can you show me your calculations because since my method is correct I think that "hidden text" answer of 3.46 is not correct. I am very sure of ≈ 3.4315 .
Last edited by bobbym (2009-06-26 05:41:26)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Lets assume that your solution is the right one, and indeed with 3.44 loads we can cross 800 miles:
with 1 load , we can go 500 miles
with 2 loads , we can go 500(1+1/3) = 666.66
with 3 loads, 500(1+1/3+1/5) = 766.66
now, it means that 3.44 - 3 = 0.44 loads we can get (800 - 766.66)x7 (we need 7 trips now to make a cache of 3 loads)
0.44 x 500miles/load = 220miles , (800 - 766.66)x7 = 233.38
0.4666 x 500miles/load = 233.3 (much closer to the answer)
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Hi uzurpatorul;
Hard to argue with that. Nice and neat proof. So if my sum is okay then the problem must be with solving it analytically and setting it equal to 800.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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if you look at my explanation is pretty much your formula for n between 3 and 4
500(1 + 1/3 + 1/5 + k/7) = 800, where k = n - 3
Another cool thing i noticed looking at your formula is that you just proof that the series is divergent, and with the common constrains any desert can be crossed (of course n grows exponentially).
P.S.
here are 2 variations of the same problem:
http://www.projecteureka.org/problem/question/105
http://www.projecteureka.org/problem/question/298
Last edited by uzurpatorul (2009-06-27 06:23:24)
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Hi uzurpatorul;
Thanks for links and the education.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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