Laying cable An island is 3.5 mi from the nearest point on astraight shoreline; that point is 8 mi from a power station (seefigure). A utility company plans to lay electrical cable underwaterfrom the island to the shore and then underground along theshore to the power station. Assume that it costs $2400/mi tolay underwater cable and $1200/mi to lay underground cable.At what point should the underwater cable meet the shore tominimize the cost of the project?IslandCable3.5 miPowerstationShore8 mi

Name: Laying cable An island is 3.5 mi from the nearest point on astraight
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Description: Laying cable An island is 3.5 mi from the nearest point on astraight shoreline; that point is 8 mi from a power station (seefigure). A utility company plans to lay electrical cable underwaterfrom the island to the shore and then underground along th

Answered: ght shoreline; that p re). A utility…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.2: Applied Problems

Problem 18E

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Laying cable An island is 3.5 mi from the nearest point on a
straight shoreline; that point is 8 mi from a power station (see
figure). A utility company plans to lay electrical cable underwater
from the island to the shore and then underground along the
shore to the power station. Assume that it costs $2400/mi to
lay underwater cable and $1200/mi to lay underground cable.
At what point should the underwater cable meet the shore to
minimize the cost of the project?
Island
Cable
3.5 mi
Power
station
Shore
8 mi

Expert Solution

Step 1

The given data is distance between shore and power station is $8 m i .$ and distance between shore and island is $3.5 m i .$

The cost to lay underwater cable is $$ 2400 / m i$ and the cost to lay underground cable is $$ 1200 / m i .$

The diagram of the above information,

Step 2

Apply the Pythagoras theorem,

$A C = \sqrt{3 . 5^{2} + x^{2}}$

$C D = 8 - x .$

To determine the point at which the underwater cable meet the shore in order to minimize the cost,

$C (x) = 2400 \sqrt{{(3.5)}^{2} + x^{2}} + 1200 (8 - x)$

Differentiating the equation,

$\frac{d C}{d x} = \frac{2400 \cdot 2 x}{2 \sqrt{3 . 5^{2} + x^{2}}} + 1200 (- 1) \frac{d C}{d x} = 0 \frac{2400 \cdot 2 x}{2 \sqrt{3 . 5^{2} + x^{2}}} + 1200 (- 1) = 0$

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