Re: Calculus
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Re: Calculus
Hi!
On Sun, 14 Jan 2001, Andrew Kolbus wrote:
> a. 200m @ $10/m, 800m @ $5/m
> b. $6000
[cut]
> 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.
Well... I used formulas, but it didn't took me more than 5 minutes. Still
interested? :-)
The idea is simple. You have two equations
c=10a+5(a+2b)
ab=60000
(a is the side next to the road and b is the other side, c is cost)
that is the same as one equation
c=900000/b+10b
the cost is minimal when dc/db=0
i.e.
-900000/b^2+10=0 and
b=sqrt(90000)=300
Then a=60000/300=200
and minimal cost is
c_min=$10*200+$5*(200+2*300)=$6000
--
Joel Kuusk E-mail: joel@scorpion.aai.ee
joelk@ut.ee
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