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Stimshieme
Pileated woodpecker
Joined: Nov 16, 2007
Posts: 176
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 Posted: Sun Oct 05, 2008 9:31 am Post subject: Math question help? Please I have an exam tommorrow! |
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DIFFERENTIATION QUESTION. PLEASE HELP. I HAVE NO CLUE ON HOW TO SOLVE THESE EXCEPT THE FIRST ONE a) i) BY SPEED = DISTANCE/TIME. THAT'S ALL I KNOW.
A ship has a 200km journey to make at a constant speed. At x km/h the cost, in $, of the journey will be:
(x^2 + 4000/x ) PER HOUR.
a)Find an expression for
i)the time taken for the journey
ii) the total cost of the trip
b). Find the speed that minimses the cost of the journey and calculate this minimum cost.
THE ANSWWERS ARE
a). i) t= 200/x
ii) 200x + 800000/x^2
b). 20 km/h, cost is $6000
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chever
'Mud'
Joined: Aug 22, 2008
Age: 20
Posts: 1668
Location: Earth
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| Posted: Sun Oct 05, 2008 12:28 pm Post subject: |
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Well the total cost of the journey is simply the cost expression multiplied by the total time, which is
[x^2 + 4000/x] * [200/x]
or 200x + 800,000/x^2
Then you want to differentiate that function in order to optimize it. That expression is
200 - 1,600,000/x^3
In this case, you can simply set that expression equal to zero in order to find the optimal value
So
200 - 1,600,000/x^3 = 0
200 = 1,600,000/x^3
200x^3 = 1,600,000
x^3 = 8,000
Then the cube root of both sides is 20.
Let me know if you don't understand anything here
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lau
Non-abelian, simple sexaginta! Male + silly bits.
Joined: Jun 18, 2006
Age: 60
Posts: 7976
Location: Somerset UK
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| Posted: Sun Oct 05, 2008 2:24 pm Post subject: |
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One little point, chever...
Having done the differentiation, and solved for a turning point of the function, you must verify that against any boundary values, etc.
In this case, the "obvious" things to check are x -> 0 and x -> inf, which both yield cost->inf, so that's OK.
The solution, x = 20, gives a cost of 6,000. which seems fine.
However, being pedantic, a speed of -inf gives a cost of -inf. I.e. don't go toward your destination, but run away from it, as fast as you can, and you will earn money.
The trouble with models, is that you always have to be sure you haven't overlooked something.
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chever
'Mud'
Joined: Aug 22, 2008
Age: 20
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| Posted: Sun Oct 05, 2008 6:33 pm Post subject: |
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Of course. That's why I said "In this case, you can simply set that expression equal to zero in order to find the optimal value"
Most optimization questions in an introductory calculus class will not throw curve balls at you like that, and this one clearly does not have such a curve ball.
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Stimshieme
Pileated woodpecker
Joined: Nov 16, 2007
Posts: 176
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| Posted: Mon Oct 06, 2008 2:35 pm Post subject: |
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SO HOW DO YOU GET 6000?
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chever
'Mud'
Joined: Aug 22, 2008
Age: 20
Posts: 1668
Location: Earth
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| Posted: Mon Oct 06, 2008 3:15 pm Post subject: |
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Plug 20 into the total cost function.
I hate to sound brusque, but you will have a lot of difficulty with introductory calculus, to say nothing of the limits / integration techniques, vector calculus and diff eq's you might take later, if you don't master these techniques.
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dark_mage
Snowy Owl
Joined: Jan 10, 2008
Age: 25
Posts: 166
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| Posted: Mon Oct 06, 2008 6:39 pm Post subject: |
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Not to mention integration is a little trickier then differentiation.
But for the problem stated by the original poster the given method works. The curve balls come in Calculus II & III (Integration & Multi-variable).
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chever
'Mud'
Joined: Aug 22, 2008
Age: 20
Posts: 1668
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| Posted: Mon Oct 06, 2008 8:36 pm Post subject: |
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I managed to get a C in Calc II.
Looks like I'm going to get an A in Calc III.
Go figure.
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dark_mage
Snowy Owl
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| Posted: Tue Oct 07, 2008 8:04 pm Post subject: |
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Yeah I got roughly the equivalent of a C in Calculus II & then obtain an AB in Calculus III (working my butt off does help though & I managed to do this without the vaunted TI-89)
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