Optimizing Boom Length for Oil Spill Containment: A Calculus Approach

sweet877
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To contain oil spills, rectangular booms that have a cross-link to provide stability are used. The cross-link joins the long sides and is parallel to the short sides. What is the minimum total length of boom required to enclose an oil spill covering 100 000 m^2 of water if it can only be constructed from 10-m sections?

I figured out how to do this if 10-m sections are not required, but cannot figure out what to do if the booms have restrictions. Please advise. Thanks!
 
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What length and width do you get if 10-m sections are not required? What are the closest multiples of 10 larger than that?
 
250 m and 400 m. But I just got that from common sense. Is there a "calculus way" to do it?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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