Calculus

[Gerald M. Carey]

Can anyone help me get started on this problem in Calculus?

A factory is located on one bank of a straight river that is 2000m
wide. On the opposite bank but 4500m downstream is a power station from
which the factory must draw its electricity. Assume that it costs three
times as much per meter to lay an underwater cable as to lay a cable over
the ground. What path should a cable connecting the power station to the
factory take to minimize the cost of laying the cable?
Thanks to all who respond

[Greg Goodknight]

Write an equation expressing total cost using only one variable. In the
general case one will travel along the bank, cross the river and continue
along the bank to the destiniation. There are reasonable simplifications to
be made, either by handwaving (perhaps involving symmetry) or by algebraic
manipulation. The amount of handwaving you can get away with is usually
proportional to how capable your teacher thinks you are.

Find a minima that suits the conditions.

Any more and I'd be interfering with your test/homework.

Cheers,

Greg
go...@nccn.net

[Anthony Hugh Back]

In article: <31B22...@email.psu.edu> "Gerald M. Carey"

<gmc...@email.psu.edu> writes:
>
> Can anyone help me get started on this problem in Calculus?
>
> A factory is located on one bank of a straight river that is 2000m
> wide. On the opposite bank but 4500m downstream is a power station from
> which the factory must draw its electricity. Assume that it costs three
> times as much per meter to lay an underwater cable as to lay a cable over
> the ground. What path should a cable connecting the power station to the
> factory take to minimize the cost of laying the cable?
> Thanks to all who respond
>
>

Let the cable go from the factory at an angle across the water to meet the
opposite bank x metres downstream. Then distance underwater will be
sqrt(2000^2 + x^2) and distance over the ground is (4500-x) metres.

Cost is 3 units under water and 1 unit over the ground, so:

Total cost C = 3*sqrt(2000^2+x^2) + (4500-x)

dC/dx = (3/2)(2x)/sqrt(2000^2+x^2) - 1 = 0 for max or min
(3/2)(2x) = sqrt(2000^2-x^2)
Square both sides 9x^2 = 2000^2 - x^2
10x^2 = 2000^2

Take sq.roots sqrt(10)*x = 2000
x = 2000/sqrt(10)
x = 632.46 metres.

So the cable meets the opposite bank 632.46 metres downstream.
--
Anthony Hugh Back