A square plot of ground 100 feet on a side has vertices labeled A, B, C, D clock

Question

A square plot of ground 100 feet on a side has vertices labeled A, B, C, D clockwise. Pipe is to be laid in a straight line from A to a point P on BC (P may be one of the vertices), and from there to point C. The cost of laying the pipe is $20 per foot if it goes through the lot (since if must be laid underground), and $10 per foot if it is laid along one of the sides of the square. What is the minimum cost for paying the pipe and how should the pipe be laid?

a) 1. Solve this problem by the Trigonometric approach.

b) 2. Solve this problem by the traditional

Explanation / Answer

Let point P be at a distance x units from B

PB = x

PC = 100 - x

AP = sqrt ( 100^2 + x^2 )

Pipe is laid along AP and CP

So, cost = 20 [ sqrt ( 100^2 + x^2 ) ] + 10 (100-x)

We wish to minimize this cost,

So, d(cost) /dx = 0

20(2x) / 2 sqrt(100^2 + x^2) = 10

So, 20x / sqrt(100^2 + x^2) = 10

sqrt(100^2 + x^2) = 2x

squaring both sides,

4x^2 = x^2 + 100^2

x = 100 /sqrt(3) = 57.73 m

Thus, total cost = 20*115.47 + 10*42.27 = 2732.10 $

Trigonometric approach:

Let angle APB = x

AP = 100 / sin x

BP = AP cos (x) = 100 / tan(x)

CP = 100 - 100/tan x

cost = 20AP +10 CP is to be minimised

f(x) = 2000 cosec(x) + 10(100 - 100cot(x) )

f'(x) = 2000 cot(x) cosec(x) - 1000 cosec^2 (x) = 0

Thus, cosec(x) / cot (x) = 2

solving we get,

cos (x) = 0.5

x = 60 degree

substituting the value of x, we get

AP = 115.47 m

CP = 100 - 100/sqrt(3) = 42.27 m

which are the values as obained earlier.

Hope this help. please rate it the best. Thanks.

Ask if you have any doubts.

Note: It is simpler to use trigonoetric approach while solving this problem. Although the answers obtained must remain the same.

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