Farmer Joe owns a cow, a goat, and a sheep. The animals each eat grass at a constant rate, and the grass grows at a constant rate. And Farmer Joe occasionally lets them eat the grass on a small pasture of his.

• If the cow and the goat graze together, the pasture is bare after 45 days.
• If the cow and the sheep graze together, the pasture is bare after 60 days.
• If the cow grazes alone, the pasture is bare after 90 days.
• If the goat and the sheep graze together, the pasture is bare after 90 days, also.

How long will it take for the pasture to be bare if all three animals graze together?

Everybody assumed that the volume of grass worked with this formula:

V'(t)=g-r, where g is the growth of the grass, and r is the eating rate for an animal.

Integrating yields:

V(t)=gt-rt+V(0), and the answer of 36 days appears after some linear algebra.

But, I think that if animals eat the grass around the edges and down to the roots so that the eaten grass cannot produce, then a formula emerges:

V'(t)=V(t)/(a+t) - r, where a is the inverse of the initial proportional growth of the grass(i.e., the initial height of the grass divided by its growth rate), and r is as before.

The differential analysis yields:

V(t)=(1+t/a)V(0)-r(a+t)ln(1+t/a)

This equation is a bit unwieldy but does show that that solution is around 36 days.  a tends to converge around 69.8 for about a 1.43% initial growth rate, with eating rates in proportion to initial volume, V(0), of .0173, .0058, and .0115.  These values produce 35.82 days for all three.

Maybe rounding error makes the answer not exactly 36.  Can anybody prove its 36 even in this scenario?

Posted by bernie on 2004-10-23 00:44:58