Calculus problem help
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Joined: May 2004
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From: Vienna, VA
first do a cross multipication to yield:
A = 28 + 70/h + 3h + 7.5
solve for dA/dh and set equal to 0
-70/h^2 + 3 = 0
3 = 70/h^2
3h^2 = 70
h^2 = 70/3 = 23.3333
h = sqrt of 23.3333 which is 4.830459
w = 28/h
w = 28/4.830459 = 5.796551
I hope I did it right as I have not done this in 10 years. I have an econ degree and did optimizations all the time.
A = 28 + 70/h + 3h + 7.5
solve for dA/dh and set equal to 0
-70/h^2 + 3 = 0
3 = 70/h^2
3h^2 = 70
h^2 = 70/3 = 23.3333
h = sqrt of 23.3333 which is 4.830459
w = 28/h
w = 28/4.830459 = 5.796551
I hope I did it right as I have not done this in 10 years. I have an econ degree and did optimizations all the time.
Joined: May 2006
Posts: 4,605
Likes: 4
From: Irvine, California
Originally Posted by magician,Mar 30 2007, 08:55 PM
Approximations! Yuck!
What's the real answer?
What's the real answer?
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Joined: Jul 2001
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From: Yorba Linda, CA
Originally Posted by steven975,Mar 30 2007, 08:57 PM
first do a cross multipication to yield:
28 + 70/h + 3h + 7.5
solve for dA/dh and set equal to 0
-70/h^2 + 3 = 0
3 = 70/h^2
3h^2 = 70
h^2 = 70/3 = 23.3333
h = sqrt of 23.3333 which is 4.83
w = 28/h
w = 28/4.83 = 5.80
28 + 70/h + 3h + 7.5
solve for dA/dh and set equal to 0
-70/h^2 + 3 = 0
3 = 70/h^2
3h^2 = 70
h^2 = 70/3 = 23.3333
h = sqrt of 23.3333 which is 4.83
w = 28/h
w = 28/4.83 = 5.80
w=sqrt(168/5)
No approximations.
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From: Irvine, California
Originally Posted by magician,Mar 30 2007, 09:09 PM
h = sqrt(70/3)
w=sqrt(168/5)
No approximations.
w=sqrt(168/5)
No approximations.
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Joined: Jul 2001
Posts: 6,592
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From: Yorba Linda, CA
Originally Posted by AlX Boi,Mar 30 2007, 10:26 PM
Is that what you require your fullerton math students to do?
Originally Posted by AlX Boi,Mar 30 2007, 10:26 PM
We left the answer to what you had in my calc classes because calculators were generally not allowed.
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From: Vienna, VA
Originally Posted by dongkeykong,Mar 30 2007, 10:50 PM
calculus! except for the derivative part (dA/dh) which was actually pretty elementary calculus.
Believe it or not this stuff has use. Say you run a factory and know how much it costs to make something, but want to make the most money possible. This is how you do it. I did this stuff in my economics coursework all the time.
Or, to a more car-related topic...
You see horsepower charts all the time. Of course you know how much peak power a car makes, but how do you know which of two horsepower curves is better? One is peaky, with a higher peak, and one is meaty with a lower peak. How do you know what is better? Well, you could do a partial integral of the relevant RPM range to find the "area under the line", which is in simpler terms the cumulative energy produced within that RPM range.
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holy sh!t! did you just say partial integral? wtf is that? j/k man. that term just vaguely reminded me of the time when i took advanced calculus back in college. all i remember from that class was that i had a french professor, 3 other classmates, and 4 questions on hour and half long exam. partial credit was my friend.