Posted 8 Mar 2013 · Report post A goat is tethered to a circular silo of radius r = 29 yd that is centered in a circular grassy field of radius R. The goat is just able to keep the silo wall clear of grass. He is also able to eat half of the grass in the annular field. How big must that field be? 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 · Report post R ~ 254.84 or sqrt{2([58Pi]^{2}-27^{2})} 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 · Report post R ~ 254.84 or sqrt{2([58Pi]^{2}-27^{2})} Smaller than the silo footprint. To clarify: 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 (edited) · Report post The goat in the picture is about 50 yards long and 40 yards tall. I would not like to meet it in a dark alley. For any rope length, a field could be built, such that the goat would have access to half of it. I guess, we are looking for the minimum size field. The minimum rope should allow the goat to reach around half of the silo. Thus the rope length is 29π yards. After that, it seems, the problem requires solving an integral. Edited 9 Mar 2013 by Prime 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 · Report post R~127 Although it's entirely possible I made a math error somewhere *shrugs*. I'm too sick/lazy to make a pretty diagram, but I divided the area the goat spans in half along the line of symmetry. The 'funky' part (i.e. where the leash is at least partially along the silo wall) can be described by the function r(theta)=29(pi-2*theta+2sin(theta)). To get the area, integrate 1/2*r(theta)^2 over theta from 0 to pi/2. Double that and subtract the area of the silo. The part where the goat's leash does not touch the silo is a semi-circle of radius 29. The rest is a little geometry, a little algebra. 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 · Report post I am wondering whether my analysis is off a bit. I get a slightly smaller R than Y-San does. But RG's answer gives exactly twice hers, meaning he may have given diameter instead of radius. So there is the possibility they are both correct (with a misplaced factor 2 in RG's case - a typo more or less.) Anyone else want to give it a shot, I'll go back over my equations meantime. Nice going, both, in any event. Prime is correct, it involves an a messy integral. And to be clear, the area the goat covers is half the annular field of grass. 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 (edited) · Report post ...actually make that 120, found a math error *whistling*...and I meant "semi-circle of radius 29*pi in that last part... Eh, the integral is not that bad...only one application of parts is needed, and a fairly simple one at that Edited 9 Mar 2013 by Yoruichi-san 0 Share this post Link to post Share on other sites

0 Posted 9 Mar 2013 · Report post If i got the question right, then R is approx. 125.54 0 Share this post Link to post Share on other sites

0 Posted 10 Mar 2013 · Report post If i got the question right, then R is approx. 125.54 dm92 your answer answer agrees with that of several others. Y-san and I get a slightly smaller number. It puzzles me (npi) what is the difference in solving this that leads to two different solutions. Maybe the wording of the OP is ambiguous. R = 121.14 yd 0 Share this post Link to post Share on other sites

0 Posted 18 Sep 2013 · Report post 121.14 seems to be correct. The area eaten by the goat is (29pi)^3/(3*29) +(29pi)^2+pi/2 = 21730.2. The field is x^2*pi=2(21730)+29^2*pi (no grass in the silo area so the radius is a little bigger) 0 Share this post Link to post Share on other sites

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A goat is tethered to a circular silo of radius

r= 29 yd that is centered in a circular grassy field of radiusR. The goat is just able to keep the silo wall clear of grass. He is also able to eat half of the grass in the annular field.How big must that field be?

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