Aptitude - Power cable question

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Power cable question

A power cable is to be run from a power plant on the bank of a river 900 meters wide to a factory that is located 3000 meters downstream on the opposite bank. If the cost of laying cable under water is Rs. 5 per meter and that of laying overhead on land is Rs. 4 per meter, find the point downstream where the cable is to cut across the river.

Total Answers and Comments: 3

Last Update: February 21, 2010 Asked by: abhijit_de2000

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Submitted by: Raze2dust

Ok..I realize this needs a bit of math(calculus, to be specific)..not sure if theres any other way out..

A<--------x------>C D
------------------------------------
^
|
900m
|
V
------------------------------------
B
<---------3000m---------------->

Let the path be A-C-B and angle BCD=t
Now CD=3000-x= 900/tan(t)=900 cot(t). So x=AB=3000-900cot(t)
BD=900
So, BC= 900/sin(t) = 900cosec(t)

Total cost = f = 4*AB+5*BC = 4(3000-900cot(t))+5(900cosec(t))
=12000-3600cot(t)+4500cosec(t)

We need to minimize f. So put df/dx=0
gives 3600cosec2(t)-4500cosec(t)cot(t)=0
3600cosec(t)=4500cot(t)..write cosec as 1/sin and cot as cos/sin..
gives cos(t)=4/5. So tan(t)=3/4.
BD/CD=3/4
900/(3000-x)=3/4
3600=9000-3x
So x=1800m

I'll try to think of a simpler method though...

Above answer was rated as good by the following members:
smokincigar07

November 09, 2008 00:58:59

Raze2dust

Member Since: November 2008 Contribution: 3

RE: Power cable question

Ok..I realize this needs a bit of math(calculus to be specific)..not sure if theres any other way out..

A<--------x------>C D
------------------------------------
^
|
900m
|
V
------------------------------------
B
<---------3000m---------------->

Let the path be A-C-B and angle BCD t
Now CD 3000-x 900/tan(t) 900 cot(t). So x AB 3000-900cot(t)
BD 900
So BC 900/sin(t) 900cosec(t)

Total cost f 4*AB+5*BC 4(3000-900cot(t))+5(900cosec(t))
12000-3600cot(t)+4500cosec(t)

We need to minimize f. So put df/dx 0
gives 3600cosec2(t)-4500cosec(t)cot(t) 0
3600cosec(t) 4500cot(t)..write cosec as 1/sin and cot as cos/sin..
gives cos(t) 4/5. So tan(t) 3/4.
BD/CD 3/4
900/(3000-x) 3/4
3600 9000-3x
So x 1800m

I'll try to think of a simpler method though...

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Overall Rating: +1

November 17, 2009 02:05:33

agkjack

Member Since: November 2009 Contribution: 1

RE: Power cable question

I do not think it is complex. They give cost to plant the cable in ground and water both also

Let assume the length across the river is x;

By given cost
5x 4*3000+5*900
so x 3300

3300/2 1650

so the point is 1650m from x axis and 450m(half of distance bet banks) from y axis
(450 1650)

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February 20, 2010 12:29:48

smokincigar07

Member Since: February 2010 Contribution: 4

RE: Power cable question

<(3000-x)->C
A------------------------------------------------------
^ |
| |
900 |
| |
| |
| |
--------------------------------------------------------
<-- x -->B

CB (900^2+x^2)^1/2

cost of laying cable on land 4*(3000-x)
cost of laying cable under water 5*(900^2+x^2)^1/2

total cost T 4*(3000-x)+5*((900^2+x^2)^1/2

to minimize T
dT/dx 0

this will give us x 1200

therefore the distance from A to C (where the cable cut though the river) is
3000-1200 1800 metres

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