A freshwater pipeline is to be run from a source on the edge of a lake to a comm

Question

A freshwater pipeline is to be run from a source on the edge of a lake to a community on an island 5 miles offshore, as shown in the figure Answer the question that follow If it costs 1.5 times as much to lay pipe in the lake as it does on land, what should x be(in miles) to minimize the total cost of this project? x = miles (Round to the nearest tenth needed.) (B) If it costs only 1.2 times as much to lay pipe in the lake as it does on land, what should x be )in mils) to minimize the total cost of this project x = miles (Round to the nearest tenth as needed.)

Explanation / Answer

Let the cost for laying pipe on land be t units of money per mile.

(A)

Then the cost for laying in water is 1.5*t unit of money per mile.

Length (in miles) of pipe on land is (10 - x) and length in water is sqrt (x^2 + 5^2)

.therefore the total cost of laying the pipes is given by

Cost(x) = t* (10 - x) + 1.5*t *sqrt(x^2 + 25)

The minimum cost is when derivative of cost function cost(x) turn 0;

ie d(cost(x))/dx=0

=>t*(-1+1.5*x/sqrt(x^2 + 25))=0

=>x/sqrt(x^2 + 25)=2/3

=>9*x^2=4*x^2+100

=>x^2=20

=>x=sqrt(20)=4.47

(B)

the cost for laying in water is 1.2*t unit of money per mile.

Length (in miles) of pipe on land is (10 - x) and length in water is sqrt (x^2 + 5^2)

.therefore the total cost of laying the pipes is given by

Cost(x) = t* (10 - x) + 1.2*t *sqrt(x^2 + 25)

The minimum cost is when derivative of cost function cost(x) turn 0;

ie d(cost(x))/dx=0

=>t*(-1+1.2*x/sqrt(x^2 + 25))=0

=>x/sqrt(x^2 + 25)=5/6

=>36*x^2=25*x^2+625

=>x^2=625/11

=>x=sqrt(625/11)=7.54

A more intelligent method would be to use Snell's law.
you can read more abt it here-http://en.wikipedia.org/wiki/Snell's_law

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