Calculus - optimizing problem

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In summary, to wire a cottage for phone service using the cheapest method, 15 meters of wire needs to be laid underwater.
  • #1
pbonnie
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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


[itex] a^2 + b^2 = c^2 [/itex]


The Attempt at a Solution


[itex] C = 15\sqrt{x^2+14400} + 10(300-x) [/itex]
[itex] C' = 15 \cdot (\sqrt{x^2+14400})' - 10 [/itex]
[itex] = 15 \cdot \frac{1}{2\cdot\sqrt{x^2+14400}} \cdot(x^2+14400)' -10 [/itex]
[itex] = \frac{15\cdot (x^2)'} {2\cdot\sqrt{x^2+14400}} -10 [/itex]
[itex] = \frac{15\cdot2x}{2\cdot\sqrt{x^2+14400}} -10 [/itex]
[itex] = \frac{15x}{\sqrt{x^2+14400}} -10 [/itex]

I'm just wondering if someone can look this over and let me know where I'm going wrong? I proceeded to solve for f'(x) = 0 and I'm getting 10.7, so I know I'm doing something wrong.

I apologize for the use of ' for prime (this was how I was taught in the lesson).

Thank you very much for your time.
 
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  • #2
Just double checking the numbers (your method looks fine), I get 107.3.

You might have typed in 1440 instead of 14400 in your calculator, which would explain the difference in exactly one order of magnitude.

Just note that this value is x, the distance not traveled above ground. The problem asks for the distance underwater so another calculation is needed after this.
 
  • #3
Oh, okay, well then I guess it was actually the next part that I wasn't sure how to do!

My attempt from there was:
[itex] 10 = \frac{15x}{\sqrt{x^2+14400}}[/itex]
[itex] 100 = \frac{15x}{x^2+14400} [/itex]

I'm going to stop there for now in case I'm already making the mistake?
 
  • #4
or if it's supposed to be
[itex] 10\cdot\sqrt{x^2+14400} = 15x [/itex]
 
  • #5
You can keep it as a fraction, but when you square each side of the equation make sure to square the numerator i.e. 15x.

Your second post is another (easier imo) way of looking at the problem. From here you can divide both sides by 10 and then square the whole problem leaving you with:

14,400+x2= 9/4 x2

14,400= 5/4 x2

11,520= x2

107.3≈ x
 
  • #6
Wow that's embarrassing.. thank you! I guess it's time for a break
 

What is the concept of optimization in calculus?

The concept of optimization in calculus involves finding the maximum or minimum value of a function. This is done by using techniques such as taking derivatives and setting them equal to zero, or using the first or second derivative test.

What is the difference between local and global optimization?

Local optimization involves finding the maximum or minimum value of a function within a specific interval or region. Global optimization, on the other hand, involves finding the maximum or minimum value of a function over its entire domain.

How do you know if a point is a maximum or minimum in an optimization problem?

In an optimization problem, a point is considered a maximum if the second derivative of the function is negative at that point, and it is considered a minimum if the second derivative is positive at that point. If the second derivative is equal to zero, further analysis is needed to determine the nature of the point.

What is the role of constraints in an optimization problem?

Constraints in an optimization problem are additional conditions or limitations that must be satisfied in order to find the optimal solution. These constraints can be in the form of equations or inequalities and help to narrow down the possible solutions.

Can calculus be used in real-life optimization problems?

Yes, calculus is widely used in real-life optimization problems in various fields such as economics, engineering, and physics. For example, it can be used to maximize profits in a business, optimize the design of a bridge, or determine the trajectory of a projectile.

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