Finding the Area or the Circumference of a gutter.

Liparulo
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Hey Guys,
For a maths assignment, we were given this question to complete:
http://nrich.maths.org/5673

I'm having an discussion with a friend of how best to approach it. Should I work out and differentiate the area or perimeter (for the circle)? If we're finding the cost of materials, then circumference would be more appropriate than the area. Or should we look at both?
Sigh, isn't it always the most basic part of a problem that stumps us?
Thanks for your assistance.
 
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You want to minimize the perimeter (or circumference) for a fixed area.
 
The same perimeter can have a varying area, correct? Or would it be the same area every time? For example, if I curve a piece of paper, I can make the shape wider or smaller, but the perimeter stays the same. Is the area changing even though perimeter stays constant? Thanks.
 
Liparulo said:
The same perimeter can have a varying area, correct?

Yes. For any given perimeter p, you can range from the minimum area of 0 - by having a rectangle with length p/2 etc. - to the maximum area of...
 
So, (sorry to continue asking questions, but this is helping), would I find the perimeter of the circumference minus the perimeter of the arc? What role does area play in determining an equation?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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