Question

A new cottage is built across the river and 300m downstream from the nearest telephone relay station. the river is 120m wide. In order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. the cost to lay wire under water is $15 per m and the cost to lay wire above the ground is $10 per m. how much wire should be laid under water to minimize cost? thank you.

Answer 1

Refer to the attachment.

Answer 2

honestly i do not understand... y dont we break it down now?

Answer 3

From the diagram, x is the length of the wire under water. 300-a is the length of wire above ground. "a" has to be calculated from the right triangle elements. Does this make sense so far?

Answer 4

Yeah it does but how comes x is slant?

Answer 5

x is slanted to represent a general x position. Remember this is a minimization problem. The possible positions for x is from straight down to directly connected to the cottage across the water.

Answer 6

ok.... to find a now, with two unknows we have square root of x^2-120^2.

Answer 7

You almost have it. You have to take the square root of x^2-120^2 because
a^2 = x^2-120^2. you want "a", not a^2.

Answer 8

x-120=a.

Answer 9

hello are you there?

Answer 10

\[a=\sqrt{x{}^{\wedge}2-120{}^{\wedge}2} \]I may have to break off. This site/my computor is mis behaving.

Answer 11

when will you come online again please. yeah like, buh if you futher simplify a also = x-120, i.e the square cancels the root. lets go futher please.

Answer 12

\[a=\sqrt{x^2-120^2} \]cannot be simplified. The cost for the wire above ground is the following:\[10\sqrt{x^2-120^2} \]

Answer 13

yeah am getting it

Answer 14

15x is the cost of the wire under water.

Answer 15

absolutely. did you multiply the cost by the sides?

Answer 16

The sum of the above ground and the below water is the total cost to install the wire.
\[15x+10\sqrt{x^2-120^2} \]

Answer 17

yeah.

Answer 18

The derivative of the total cost is:\[15+\frac{10 x}{\sqrt{-14400+x^2}} \]

Answer 19

Combine the fractions above:\[\frac{5 \left(2 x+3 \sqrt{-14400+x^2}\right)}{\sqrt{-14400+x^2}} \]

Answer 20

combine wat fraction?

Answer 21

Combine the terms may be a more accurate statement. the result is the following:\[\frac{5 \left(2 x+3 \sqrt{-14400+x^2}\right)}{\sqrt{-14400+x^2}} \]

Answer 22

Solve \[2 x+3 \sqrt{-14400+x^2}=0 \]for x.

Answer 23

\[\left(3 \sqrt{-14400+x^2}\right)^2=(-2 x)^2 \]\[9 \left(-14400+x^2\right)=4 x^2 \]\[x=\pm 72 \sqrt{5} \]

Answer 24

you got me confused now.......where did you get 2x +3

Answer 25

what is the derivative of the total cost.

Answer 26

\[15+\frac{10 x}{\sqrt{-14400+x^2}} \]

Answer 27

the derivative of 10 is 1.

Answer 28

Total cost is\[15x+10\sqrt{x^2-120^2} \]The derivative of the above is\[15+\frac{10 x}{\sqrt{-14400+x^2}} \]Solve the following for x\[15+\frac{10 x}{\sqrt{-14400+x^2}}=0 \]

Answer 29

Sorry, but I have to break off now.

Answer 30

am still trying to get it right. thanks bro.

Answer 31

ill get back to you. thanks alot. you are a life saver.