🚀 go-pugleaf

David Ullrich (ullrich@math.okstate.edu) wrote:
: molinari@math.lsa.umich.edu (Rory Molinari) wrote:
: >
: > >>>>> "James" == James Xidos <jamesx@morgan.ucs.mun.ca> writes:
: > In article <3kkpsv$m60@coranto.ucs.mun.ca> jamesx@morgan.ucs.mun.ca (James Xidos) writes:
: >
: > [snip]
: >
: > >>>>  <tealsl@fox.nstn.ca> writes:
: > >>>>
: > >>>> > " A line goes through the point (2,2). It cuts the x axes at A
: > >>>> > and the y axes at B. What is the minimum length of AB"
: > >>>>
: >
: > [snipped:  discussion of why answer is zero]
: >
: > James>    Many problems in mathematics have the TRIVIAL solution of
: > James> zero.  Considering that question requires the student to use
: > James> calculus to answer it, perhaps the non-zero solution is
: > James> required. I would suggest getting the A-B distance as a
: > James> function of some parameter, then getting the 1st derivative
: > James> with respect to this parameter to solve for the minimum.
: >
: > [snip]
: >
: > In this problem we can get _any_ non-negative length for AB, so
: > zero (although possibly trivial) is not an isolated answer, and
: > we're not going to get anything non-zero by minimizing.

: 	Having made an ass of myself once already by overlooking the
: trivial solution I feel uniquely qualified to say "Oh yeah? What line
: gives you length(AB)=1?"
: 	The minimum is zero but it's isolated - the next-largest
: possibility is 4*sqrt(2).
: Dave Ullrich.

For heavens sake draw a picture. Or else do this:
Find the distance AB as a function of slope for all possible lines
thru (2,2). This is well defined function for all slopes except
m=0, which will be an asymptote. Now take the derivative. You'll
get two critical points at m=1 (the line y=x) and at
m=-1 (The line y=4-x).  The first is the absolute min, the second a
relative min.

Sheeesh!