9 replies to this topic

[bonanova]

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?

[Rob_Gandy]

Spoiler for I got

R ~ 254.84 or sqrt{2([58Pi]²-27²)}

[bonanova]

Spoiler for I got

R ~ 254.84 or sqrt{2([58Pi]²-27²)}

 

Smaller than the silo footprint.

To clarify:

[Prime]

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.

 

Spoiler for seems like calculus

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 by Prime, 09 March 2013 - 06:15 AM.

[Yoruichi-san]

Spoiler for I get...

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.

[bonanova]

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.

[Yoruichi-san]

Spoiler for Yeah...

...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 by Yoruichi-san, 09 March 2013 - 09:27 AM.

[bonanova]

Nice going.

[dark_magician_92]

If i got the question right, then

Spoiler for

R is approx. 125.54

 

[bonanova]

If i got the question right, then

Spoiler for

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.

 

Spoiler for For the record I get

R = 121.14 yd