Re: Calculus
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Re: Calculus
a.
200m @ $10/m, 800m @ $5/m
b.
$6000
I am
not sure if this is correct, I just used a 200 x 300 m field (200m along the
road). I am kind of tired as it is 11:30 pm right now. Let me know
if anyone gets a different answer.
If
anyone is wondering how I came up with 200 x 300, there was really not a lot of
calculation behind it, which is why it is probably wrong. To make things
easy, I decided the length along the road was the length (L), and the other
dimension was the width (W). I figured the cost for each meter of L
was $15 ($10 along the road + $5 for the opposite side), and the cost for each
meter of W was $10 ($5 for each side). I saw the ratio of cost in W:L as
3:2, so I picked a set of factors of 60 000 with the same ratio (I figured that
making the cost of each dimension the same would make it like a square, and a
square is the quadrilateral with the greatest area to perimeter ratio, so I
chose W=200 and L=300).
My
logic is probably a bit flawed, but if someone actually figures it out using
formulas or something (read: anything taking longer than 5 minutes =) ), let me
know. I am interested in the solution found by alternate
means.
Andrew
Kolbus
Check this problem out people. Tried almost everything...? I am perplexed
by it, see what you can do with it- have fun.
---
A farmer wants to fence in 60 000 mē of land in a rectangular field
along a straight road. The fencing that he plans to use along the road costs
$10 per metre and the fencing that he plans to use for the other three sides
costs $ 5 per metre.
a.-how much of each type of fence should he buy to keep expenses to
a minimum
b.-what is the minimum expense
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References:
- Calculus
- From: Arthur Labenek <labenek@HOTMAIL.COM>