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 Posted: Sun Jul 24, 2011 2:17 am
I'm not sure if I did this right, this has been driving me up a wall for hours. My brains hurts. sad 
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Posted: Sun Jul 24, 2011 2:49 am
First off, I don't see how your 2 magically became a 4 in the second line of the Area calculations.
Second, I expect that you will have the value of pi somewhere in your answer, so don't get worried if it turns out to be that way.
Third, if you take the 'r' value that you have received, and plug it back into the area equation, you get a ridiculous answer that is most likely false.
Follow these and you should get a better answer. I can't do all of the computation right now, considering my current mental state, but hopefully this helps.
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Posted: Tue Jul 26, 2011 5:20 pm
Total perimeter: 2x + 2πr = 400
Rectangular area: Α = r(2x) = r(400-2πr)
So your answer of r = 100/π as your critical point is correct.
The only thing left to do is actually plug it into the area: A = (100/π)(400 - 200) = (20000/π) m².
@Sioga: derivative of r is 1, but derivative of r² is 2r. That's where the 4 comes from.
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Posted: Tue Jul 26, 2011 8:47 pm
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Posted: Wed Jul 27, 2011 2:03 am
Though technically, we don't even have to treat it as a calculus problem. Since A = r(400-2πr) is quadratic, let's complete the square:
-A/2π = r² + (200/π)r = (r - 100/π)² - 10000/π²
A = -2π(r - 100/π)² + 20000/π
The vertex is at r = 100/π with value A = 20000/π. We know it's a maximum and not a minimum because the parabola opens downward (sign of leading coefficient).
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