Minimising the materials used for a circular gutter.

Liparulo
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Homework Statement


A plastic gutter is designed to catch water at the edge of a roof.


Manufacturers need to minimise the amount of material used to make their product. What is the best cross-section for a gutter? In this case, a circle.


Homework Equations


None, asside from Geometry.


The Attempt at a Solution


From my thinking, we need to minimise the perimeter of the circle (minus the gap for water to be let in - the arc length) and maximise the area. My problem is, how do I go about it? We would need to find the circumference minus the arc length and differentiate that to find the minimum perimeter and then set it against the maximised area of the circle? I'm rather confused. :|
 
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There really isn't enough given information to fully solve the problem, the way I see it.
 
Well, here's the problem in its entirety:
http://nrich.maths.org/5673

We weren't given anything else to work with.
 
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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