Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.
|
|
|
|
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(T-B(m+1)ny) / (A(n+1))
F'(y) = (-n)(2B(m+1)ny-T) / (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 mother-of-all-fence-problems:
>>
>> 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 east-west fence runs
>> and n+1 north-south runs.) Suppose that fencing costs $A/foot
>> for east-west runs and $B/foot for north-south 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 east-west fencing
>> and $T/2 on north-south 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.
|
|
|
|
