Finding the derivative in an optimization problem

In summary, the question asks for the minimum amount of wire that should be laid under water to connect a cottage to a telephone relay station on the other side of a 120m wide river. The cost of laying wire is $15/m under water and $10/m above ground. Using the Pythagorean theorem, an overall equation is formed to represent the costs involved. However, the use of the quotient rule is unnecessary and a more precise description of the variables is needed. Multiplying the cost to the distance may be a better approach to finding the derivative.
  • #1
mmmhmmk
4
0
Hi all, I've been stuck on this question for hours and hours, I'm not sure what I'm doing wrong..

The question states,

"a new cottage is built across the river and 300 m downstream from the nearest telephone relay station. The river is 120 m wide. 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 costs to lay wire are:
-under water: $15/m
-above ground: $10/m
How much wire should be laid under water to minimize cost?"This is what I got so far:

distance of wire above ground = (300-x)/10
distance of wire under water = sqrt(x^2+120^2)/15
x= distance between the point directly across the cottage and the point where the wire crosses the river to the other side.
sqrt(x^2+120^2) = distance from the cottage to the other side of the river, aka the route of the wire to be laid under water. this is a right angle triangle so the pyth. theorem can be used to find out the hypotenuse.The overall equation:

C= sqrt(x^2+120^2)/15 + (300-x)/10

because the problem asks for the cost, so the costs are in the denominators and distances in the numerators.Now.. I'm not sure if i use the quotient rule or not, but I'm stuck differentiating the top half of the underwater fraction..

I've tried putting the two fractions together using a common denominator, and THEN using the quotient rule, but i still don't understand how to differentiate the top - it just makes it even more confusing.
doing that, I got:

C`= [ ( 10(x^2+120^2)^0.5 + 15(300-x) )`(150) + 150`( 10(x^2+120^2)^0.5 + 15(300-x) ) ] / 150^2

C`= ( 10(x^2+120^2)^0.5 + 15(300-x) )`(150) / 22500

i'm not sure what to do next, or if i should even do it this way..
if anybody could explain it please, you could even use a different example that's similar, that would be great
i just don't understand the whole idea..

thanks a lot
 
Last edited:
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  • #2
mmmhmmk said:
Hi all, I've been stuck on this question for hours and hours, I'm not sure what I'm doing wrong..

The question states,

"a new cottage is built across the river and 300 m downstream from the nearest telephone relay station. The river is 120 m wide. 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 costs to lay wire are:
-under water: $15/m
-above ground: $10/m
How much wire should be laid under water to minimize cost?"





This is what I got so far:

above ground = (300-x)/10
under water = sqrt(x^2+120^2)/15
What does "above ground" represent? How can "above ground" be equal to a number? Same comments about "under water". Try to be more descriptive and more precise about what the expressions represent. Think about what you're trying to minimize.
mmmhmmk said:
The overall equation:

C= sqrt(x^2+120^2)/15 + (300-x)/10

Now.. I'm not sure if i use the quotient rule or not, but I'm stuck differentiating the top half of the underwater fraction..
You shouldn't be using the quotient rule at all in this problem.
mmmhmmk said:
I've tried putting the two fractions together using a common denominator, and THEN using the quotient rule, but i still don't understand how to differentiate the top - it just makes it even more confusing.
doing that, I got:

C`= [ ( 10(x^2+120^2)^0.5 + 15(300-x) )`(150) + 150`( 10(x^2+120^2)^0.5 + 15(300-x) ) ] / 150^2

C`= ( 10(x^2+120^2)^0.5 + 15(300-x) )`(150) / 22500

i'm not sure what to do next, or if i should even do it this way..
if anybody could explain it please, you could even use a different example that's similar, that would be great
i just don't understand the whole idea..

thanks a lot
 
  • #3
Mark44 said:
What does "above ground" represent? How can "above ground" be equal to a number? Same comments about "under water". Try to be more descriptive and more precise about what the expressions represent. Think about what you're trying to minimize.
You shouldn't be using the quotient rule at all in this problem.


above ground represents the wire that will be laid above ground on the other side of the river, which costs $10/m
under water represents the wire that will be laid under water, across the river, costs $15/m


What rule should I use for the equation?
 
  • #4
How does "above ground" represent the wire -- its color, its gauge, its weight, what? Your description "above ground" is extremely vague. I'm trying to get you to think more clearly.
 
  • #5
mmmhmmk said:
above ground represents the wire that will be laid above ground on the other side of the river, which costs $10/m
under water represents the wire that will be laid under water, across the river, costs $15/m


What rule should I use for the equation?
No, they don't. "wire" is not a number so what you are calling "above ground" CAN'T be "the wire ... ". I think you mean "the distance the wire goes above ground ..."- say that.
You also start writing equations, (300- x)/10, etc. without saying what "x" represents!
 
  • #6
Just curious where you came up with those equations. I would have used very different ones.
 
  • #7
yup, the distance is what i meant to say, hallsofivy.

munaught - there was a similar question in my textbook, just with different numbers. I followed the steps they used for that one and got stuck trying to figure out the derivative, because the textbook does not explain that step.
and since I'm trying to figure out the cost, the cost of each distance of wire is in the denominator

here is a picture of the problem, so there is no more confusion.

http://tinypic.com/r/1zlxafo/3

could anyone please help me figure out what to do? and tell me what I'm doing wrong? I'm still not sure what rule to use to differentiate

you don't have to give me the answer, just please, walk me through getting the derivative of the equation

thank you.edit: sorry,pic didnt show up, here's the link. http://tinypic.com/r/1zlxafo/3
 
Last edited:
  • #8
would multiplying the cost to the distance work, instead of dividing?

Cost = 15sqrt(x^2+120^2) + 10(300-x)

i still need help with the derivative

C`= (15)(0.5)(x^2+120^2)^-0.5 + (10)(300-x)`
 
  • #9
Yes, you are looking for the distance and then multiplying by the cost per meter to get the total cost, not dividing.

Your equation should be set up as:

C = 15(sqrt(x2+1202) + 10(300-x)

So differentiate with respect to x, then set the derivative equal to zero and solve for x.

I won't give you an exact answer but I got a value of 100<x<120 which is perfectly reasonable. Once you have x though, remember to plug it back into the distance across the stream to get your final answer.
 

1. What is the derivative in an optimization problem?

The derivative in an optimization problem is the rate of change of a function with respect to its independent variable. In other words, it measures how much the output of a function changes when the input is changed by a small amount.

2. Why is finding the derivative important in an optimization problem?

Finding the derivative is important in an optimization problem because it allows us to determine the critical points of a function, where the derivative is equal to zero. These critical points are often the points at which a function reaches its maximum or minimum value, making it essential for solving optimization problems.

3. How is the derivative used to solve optimization problems?

The derivative is used to solve optimization problems by setting it equal to zero and solving for the critical points. These critical points can then be plugged back into the original function to determine the maximum or minimum value.

4. Can the derivative be used to find the optimal solution in all optimization problems?

No, the derivative cannot always be used to find the optimal solution in all optimization problems. In some cases, the derivative may not exist or may be equal to zero at multiple points, making it difficult to determine the optimal solution. Other methods, such as setting up and solving equations, may be needed in these cases.

5. Are there any limitations to using the derivative in optimization problems?

Yes, there are limitations to using the derivative in optimization problems. For example, the derivative may not exist for certain functions or may be too complex to calculate. In these cases, alternative methods may need to be used to solve the optimization problem.

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