Optimizing Wire Usage for Telephone Service Installation

  • Thread starter the_morbidus
  • Start date
  • Tags
    Calculus
In summary, the question asks how much wire should be laid under water to minimize cost for wiring a cottage for phone service. The cost for laying wire under water is $15 per meter and the cost for laying wire above ground is $10 per meter. By using Pythagoras' theorem, the length of wire above ground is 300-x and the total cost function is P=1/15 (x^2+14400)^1/2 +1/10 (300-x). To find the minimum cost, the derivative of the function is taken, but a mistake in the approach results in a negative answer. The correct approach would be to set the cost of laying a meter of cable underwater as $15, making it more expensive
  • #1
the_morbidus
18
0

Homework Statement


A new cottage is built across the river and 300 m 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 ground is $10 per m. How much wire should be laid under water to minimize cost?


Homework Equations


Pythagoras theorem a^2 +b^2 = c^2


The Attempt at a Solution



ok so i think the wire going across the river will create a triangle

|\...| ^
|..\...| |
|...\...| x
|...\..| |
|...\| V
|...|
|...| 300-x
|_120 _| won't allow me to make a proper diagram but hope you guys get the picture so try to imagine it without all the dots on it.

so the equation for the hypotenuse for the length of wire under water will be sqrt(x^2 +120^2) = sqrt(x^2 +14400)

the equation for the length of wire above ground is 300-x

the function for the price for the total wire usage is P= sqrt(x^2 +14400)/15 + (300-x)/10
P=1/15 (x^2+14400)^1/2 + 1/10 (300-x)
now for the derivative of the function

P'=1/15 *1/2 (x^2 +14000)^-1/2 *2x +1/10(-1)
P'=x/(15(x^2 +14400)^1/2) - 1/10
now to find the value for x when the function equals 0

x/(15(x^2 +14400)^1/2) - 1/10=0
x/15(x^2 +14400)^1/2 = 1/10
10x=15(x^2+14400)^1/2
(10x)^2=(15(x^2+14400)^1/2)^2
100x^2=225(x^2+14400)
100x^2=225x^2 +3240000
-125x^2=3240000
x^2=3240000/-125
x^2=-25920
x=sqrt(-25920) see the problem is the negative answer so i know i have made a mistake on my approach, because i was thinking once i have obtained the value of x i could add it so the equation sqrt(x^2 +14400) and i would obtain the minimum amount of wire laid under water to minimize the cost.

Can anyone help me?? thanks!
 
Last edited:
Physics news on Phys.org
  • #2
If the length underwater is L meters, then the cost of laying the cable underwater is 15*L, not L/15. You've set it up so the cost of laying a meter of cable underwater is $(1/15). That makes it cheaper to lay the cable underwater than to string in on land, so the cheapest way is to make it ALL underwater. That's why you aren't getting a critical point in a realistic place.
 
  • #3
o i see, my mistake, thanks for your advice !
 

1. What is calculus?

Calculus is a branch of mathematics that deals with the study of change and motion. It involves using mathematical models and techniques to solve problems related to rates of change, optimization, and integration.

2. How is calculus used in real life?

Calculus is used in a wide range of fields, such as physics, engineering, economics, and statistics. It is used to solve problems involving rates of change, optimization, and determining areas and volumes of irregular shapes.

3. What are the two main branches of calculus?

The two main branches of calculus are differential calculus and integral calculus. Differential calculus deals with rates of change and slopes, while integral calculus deals with areas and volumes.

4. What are some common applications of calculus?

Calculus has many practical applications, such as determining the maximum profit for a business, optimizing the design of a bridge, predicting the motion of planets, and analyzing the spread of diseases.

5. Is calculus difficult to learn?

Calculus can be challenging for some people, as it involves complex mathematical concepts and techniques. However, with practice and a solid understanding of the fundamentals, it can be learned and applied effectively.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
745
  • Calculus and Beyond Homework Help
Replies
22
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
3
Views
2K
  • Math Proof Training and Practice
3
Replies
100
Views
7K
  • Math Proof Training and Practice
3
Replies
102
Views
7K
Back
Top