krs@amdahl.UUCP (Kris Stephens) (03/29/85)
A friend of mine has used the following as an Interview Question:
A farmer has a fenced 10-meter by 20-meter field the occupant
of which is a "point cow" {see footnote}. When tethering the
cow to a corner-post of the field, how long does the rope need
to be so that the cow can eat half of the grass in the field?
- - - - - -
Note: A "point cow" is a relative of the famous "frictionless
puck", whose [USDA :-)] prime attribute is that its head
occupies the point at the end of its tether (a cow with
a null-size head and neck).
This is a deceptive problem. The only answers I've seen were from
Engineers, who offered to tell me the answer "to as many decimal places
as I want" by successive approximation, which, as we all know, is _not_
the answer.
My friend's reasoning in posing the question is to watch how the person
being interviewed approaches problem-solving, rather than actually looking
for the answer.
--
Kris Stephens (408-746-6047) {whatever}!amdahl!krs
[The opinions expressed above are mine, solely, and do not ]
[necessarily reflect the opinions or policies of Amdahl Corp. ]tallman@dspo.UUCP (04/02/85)
> A farmer has a fenced 10-meter by 20-meter field the occupant > of which is a "point cow" {see footnote}. When tethering the > cow to a corner-post of the field, how long does the rope need > to be so that the cow can eat half of the grass in the field? > Kris Stephens (408-746-6047) {whatever}!amdahl!krs I can see why this is a good interview question - it is tempting to try to integrate the "slice out of a circle" shape, which gets very messy. An easier way is to divide the shape into a circular wedge and a triangle. ----------------- - - The o's mark the circular arc - - of the rope's maximum extent. - - The *'s mark a line from the -ooo - tether corner to the point of 2x - oo - intersection of the arc with - o - the fence. - o - Let r be the rope length, x - r * - the length of the short side (10m), - * - and phi the angle marked. -* phi - 0 < phi < pi/4. ----------------- x phi = arccos(x/r) r = x/cos(r) area of triangle = x*x*tan(phi)/2 area of circular wedge = pi*r^2*(1/4 - phi/(2*pi)) [fraction of a circle] desired area = x*(2*x)/2 = x^2 x^2 = x^2*tan(phi)/2 + pi*(x^2/cos^2(phi))*(1/4 - phi/(2*pi)) 1 = tan(phi)/2 +(pi/cos^2(phi))*(1/4 - phi/(2*pi)) cos^2(phi) = sin(phi)*cos(phi)/2 + pi*(1/4 - phi/(2*pi)) phi = sin(phi)*cos(phi) - 2*cos^2(phi) + pi/2 phi = sin(2*phi)/2 - (cos(2*phi) + 1) + pi/2 let theta = 2*phi, 0 < theta < pi/2 theta/2 = sin(theta)/2 - cos(theta) - 1 + pi/2 theta = sin(theta) - 2*cos(theta) - 2 + pi > This is a deceptive problem. The only answers I've seen were from > Engineers, who offered to tell me the answer "to as many decimal places > as I want" by successive approximation, which, as we all know, is _not_ > the answer. I have no problem with an honest numerical approximation if the solution can't be put in closed form (and very few real-world problems can be). By Newton's method I found: theta = 1.07927646 radians r = 10/cos(theta/2) = 11.646443 meters Perhaps the problem is a trick question because it does not say the cow must only reach exactly half the field. In this case, any length greater than this solution will do. Say 25 meters, for example. C. David Tallman - dspo!tallman@LANL or ucbvax!unmvax!lanl!dspo!tallman Los Alamos National Laboratory - E-10/Data Systems Los Alamos, New Mexico - (505) 667-8495 -- C. David Tallman - dspo!tallman@LANL or ucbvax!unmvax!lanl!dspo!tallman Los Alamos National Laboratory - E-10/Data Systems Los Alamos, New Mexico - (505) 667-8495