Question

Please help with this calculus problem!



Two cottages without electric services are situated 20m and 40m from a buried electric power line. Lines drawn from the cottage that meet the power line at right lines are 30m apart.

i) Where should the junction box be located on the power line to minimize the length of the connection cable to the cottages?
For this I got:
2x^2-60x+2900
dy/dx 4x-60=0
x=15

ii) Are there other ways to connect the cottages to the power line? Discuss the merits of these alternatives.

Picture can be seen here: http://tupper.vsb.bc.ca/math/text/p403.pdf

Answer 1

Can anyone please help me?

Answer 2

I really need help with part ii), I think I did part i) correctly.

Answer 3

let it be x

Answer 4

Let what be x?

Answer 5

the distance ...
and the other side is going to be 30-x

Answer 6

then you have distances = (20^2+x^2)^(1/2) + (40^2+(30-x)^2)^(1/2)
now find the critical points, test for minimum, then you will have the value of x,

Answer 7

from wolfram calculator you have x=10 as your answer,
http://www.wolframalpha.com/input/?i=extremum+%2820%5E2%2Bx%5E2%29%5E%281%2F2%29+%2B+%2840%5E2%2B%2830-x%29%5E2%29%5E%281%2F2%29
I leave the rest of the steps to you

Answer 8

Yeah, for the first part I got,
\[20^{2}+x ^{2}=d _{1}^{2}\]
\[40^{2}+(30-x)^{2}=d _{2}^{2}\]
\[=20^{2}+x ^{2}+40^{2}+x ^{2}-60x+900\]
\[=2x ^{2}-60x+2900\]
\[dy/dx = 4x-60=0\]
\[x=15\]

Answer 9

Where do you get the from 1/2 from?

Answer 10

distance formula ... you forgot to add 1/2 power

Answer 11

So the answer isn't x=15? By the way, those are the steps that I took for part i).

Answer 12

ii) connect cottage that is 40m away, with a line to cottage that is 20 m away, and cottage that is 20m away to the power line. total wire length is +-36m +20m = 56m which is shorter. shorter wire is an advantage. This assuming the cottages lie on the same side of the power line.

Answer 13

Where do you get 36m from?

Answer 14

do you now distance formula?? right??

Answer 15

No...

Answer 16

do you know pythagoras theorem??

Answer 17

yeah, a^2 + b^2 = c^2

Answer 18

do you think squaring will give distance??

Answer 19

I'm confused...

Answer 20

you are trying to get the distance by square of d
d^2 will not give you distance.
d will give distance

add those two distances d1 and d2

Answer 21

=20^2+x^2+40^2+x^2−60x+900
This is d1^2 + d2^2

Answer 22

no d1+d2

Answer 23

Yea so you woul;d have to sqrt 20^2+x^2+40^2+x^2−60x+900 right?

Answer 24

yeah that will be your function
you know how to find minima right??

Answer 25

No... Ok I have a question. Do I sqrt 20^2+x^2+40^2+x^2−60x+900 or 20^2+x^2+40^2+(30-x)^2?

Answer 26

it's the same thing ...i guess.

Answer 27

So I would get 40 + 20 + 30 + x - x
= 90?

Answer 28

do you know how to take derivative??

Answer 29

Yes

Answer 30

you know how to find critical points??

Answer 31

Nope.

Answer 32

take the first derivative ... equate it to zero, and find the values of x.
the value of x will be your critical points
and please refer her
http://answers.yahoo.com/question/index?qid=20080127144920AA6bcrW

Answer 33

*here

Answer 34

The first derivative of 90?

Answer 35

Hello?

Answer 36

yeah ..

Answer 37

Wouldnt that be 0?

Answer 38

i didn't get you?? sorry ... first derivative of 90 from where??

Answer 39

"So I would get 40 + 20 + 30 + x - x
= 90?"

Answer 40

where??

Answer 41

Scroll up

Answer 42

oh this again, still didn't get it?

Answer 43

Nope :P

Answer 44

@TuringTest
I think he still doesn't know how to find the extremum

Answer 45

good point, do you know how to find an extremum @JoBo ?

Answer 46

I do now

Answer 47

and the problem above is quite complicated one.

Answer 48

But Im not very good at it :/

Answer 49

can I prove it for the general case starting from scratch?
that is. I can prove that you always put the box in the middle no matter how high the houses are or how far apart they are
otherwise I might get bored...

Answer 50

LOL, yeah sure go for it

Answer 51

ok, let's draw two houses ant some distance apart \(R\) and some height \(h_1\) nad \(h_2\) respectively

Answer 52

Ok

Answer 53

now we want to connect a power line to the ground at some point between the houses at a point \(x\)