Finding the Optimal Leash Length for a Spherical Grazing Area

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The discussion revolves around a mathematical problem involving a spherical volume of jelly surrounded by a spherical cage, where the goal is to determine the length of a leash that allows an animal to consume half of the jelly's volume. The volume of a sphere is noted, and the proposed solution suggests that the leash length should be between the radius (r) and twice the radius (2r). The poster attempts to solve the problem using calculus, specifically by integrating the volume swept out by the leash and equating it to half the volume of the sphere. After setting up the equations and solving, they arrive at a numerical value of approximately 1.2285 for the leash length when the radius is normalized to 1. However, they express uncertainty about the validity of this solution when the radius is not equal to one, questioning whether the ratio derived would hold true for different radii.
JCienfuegos
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Hello everyone
This problem is similar to a problem that appeared sometime back on this website called "last geometry challenge, very difficult!"
The problem was this: There is a circular field of grass of radius r surrounded by a fence. If aa sheep is tethered to the fence, how long should its leash be so that it eats only 1/2 of the grass in the circle.
I propose making this into a 3D problem. Make the circle of grass a sphere of jelly surrounded by a spherical metal cage. How long should the leash be if an animal is tethered to the surrounding spherical metal cage so that it eats only 1/2 of the volume of jelly.
Ill post the answer in a couple of days if anyone is interested.
Julian
 
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Since the volune of a sphere is
V = (4 pi ) )r^44 / 3

Then the answer is somewhere between r and 2r.
( a range is an answer ) :)
 
I tried using this trick in high school. The teacher said what is sin342 or something like that and I put "something between -1 and 1". I got that question wrong.
 
I now get the leash to be 1/10 longer than the radius of the sphere.
Is that correct?
 
Well, that is not the answer i got. Now that think about it, my solution may be wrong, but I cannot think why.
What I did was this:
I imagined the sphere to be resting onto of the x axis, directly on top of the origin. I then tied the leash to the origin. I found the equation of the sphere would be the circle x^2 + (y-r)^2 = r^2 rotated around the y axis. The volume swept out by the leash would be the circle x^2 + y^2 = l^2. I revolved the region bounded by these two curves around the y-axis and set it equal to 1/2 * 4/3 * pi r^3. I assumed the radius to equal 1, since I figured the units of its length wouldn't change the answer.
Then I set up some integrals, and solved. I am left with 8l^3-3l^4-8=0, which I solved numerically and I got 1.2285.
The trouble is, what is the radius isn't one? Then the ratio 1:1.2285 isn't the same. If it is okay for the radius to be simply 1, then the answer is good.
 

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