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Topic:
Re: word problems and ``relevance'' CORRECTION
Replies:
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Re: word problems and ``relevance'' CORRECTION
Posted:
May 23, 2001 3:59 PM



Whoops, Right idea but some wrong equations. Correct solution:
Let x and y be the length and width of one of the m*n pens.
The over all dimensions are then mx by ny
The cost in one direction is Axm(n + 1) and in the other is Bny(m + 1)
and Axm(n + 1) + Bny(m + 1) = T
so x = (T  B(m + 1)ny) / (Am(n + 1))
The total fenced area F is mx*ny and F(y) = ny(TB(m+1)ny) / (A(n+1))
F'(y) = (n)(2B(m+1)nyT) / (A(n+1)) and F'(y) = 0 when y = T / (2B(m+1)n)
Substituting into Axm(n + 1) + Byn(m + 1) = T gives
Axm(n + 1) + T/2 = T
or Axm(n + 1) = T/2 and thus Bny(m + 1) = T/2
QED
Sorry
Lin
In a message dated 5/23/2001 3:09:45 PM Eastern Daylight Time, LnMcmullin@AOL.COM writes:
> In a message dated 5/22/2001 5:19:39 PM Eastern Daylight Time, > zorn@STOLAF.EDU writes: > > > > >> PS Here's a candidate for motherofallfenceproblems: >> >> Farmer John plans a rectangular, m by n ``fence grid'' , with >> m*n ``pens'' in all. (John could arrange this  assuming >> there aren't any convenient rivers or barns to substitute for >> some of the fencing  by erecting m+1 eastwest fence runs >> and n+1 northsouth runs.) Suppose that fencing costs $A/foot >> for eastwest runs and $B/foot for northsouth runs. (Yes, this >> does sounds absurd, but maybe John farms on steeply sloped land.) >> Suppose also that John has $T to spend, and that he >> wants the largest *total* penned area. >> >> Show that John should spend half his money ``in each direction.'' >> (In other words, John should spend $T/2 on eastwest fencing >> and $T/2 on northsouth fencing.) Does the answer change if >> one or more sides is a river? >> >> > > > Paul, > > Nice problem! > > Let x and y be the length and width of one of the m*n pens. > > The over all dimensions are then (m + 1)x by (n + 1)y > > The cost in one direction is A(m + 1)x and in the other is B(n + 1)y > > and A(m + 1)x + B(n + 1)y = T > > so x = (t  B(n + 1)y) / (A(m + 1)) > > The total fenced area F is (m + 1)x * (n + 1)y and F(y) = (1/A)*((n + 1)(T >  > B(n + 1)y)y) > > F'(y) = (1/A)(n + 1)(2B(n + 1)y  T) and F'(y) = 0 when y = T/(2B(n + 1)) > > Substituting into A(m + 1)x + B(n + 1)y = T gives > > A(m + 1)x + T/2 = T > > or A(m + 1)x = T/2 and thus B(n + 1)y = T/2 > > QED > > Yes, I used my TI92+ > > Lin
Lin McMullin Niantic, CT.



