Stuck with Optimisation question, help?

In summary: Let someone else handle it.In summary, the conversation revolves around finding the maximum value for the speed of a signal along a communications cable with a copper core and insulating sheath, using the given equation S=8x^2ln(1/2x). The conversation includes discussions on differentiation, the use of the product rule and chain rule, and the suggestion of using second derivative test to find the maximum. However, there is confusion on whether partial derivatives are needed and the problem being a single or multivariable problem.
  • #1
CallumC
20
0
S=8x2ln(1/2x)

Find the value of x that gives a maximum.

So far I have got, by differentiating: x2+ln(1/2x). [could be wrong]

Btw the way in the question x is a ratio and so cannot equal zero.

Please help and explain how to do it thanks :)
 
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  • #2
First, brush up on taking derivatives. What you have in the OP is way wrong.
 
  • #3
Can you show the steps you took when you differentiated the function?
 
  • #4
SteamKing said:
First, brush up on taking derivatives. What you have in the OP is way wrong.

I'll take you through my working:

[itex]\frac{dS}{dx}[/itex]=16x3+16xln(1/2x)

16x3+16xln(1/2x)=0. For maximum

16x(x2+ln(1/2x))=0

X2+ln(1/2x)=0
 
  • #5
Number Nine said:
Can you show the steps you took when you differentiated the function?

I'll take you through my working:

[itex]\frac{dS}{dx}[/itex]=16x3+16xln(1/2x)

16x3+16xln(1/2x)=0. For maximum

16x(x2+ln(1/2x))=0

X2+ln(1/2x)=0
 
  • #6
I'm confused. In the OP, S = 8x^2 * ln(1/2x)

In post #4, dS/dx = 16x^3 + 16x*ln(1/2x)

?

I think it would be better if you post the entire problem from the beginning and don't try making shortcuts.
 
  • #7
yeah the power rule bud, did you use it right?
 
  • #8
SteamKing said:
I'm confused. In the OP, S = 8x^2 * ln(1/2x)

In post #4, dS/dx = 16x^3 + 16x*ln(1/2x)

?

I think it would be better if you post the entire problem from the beginning and don't try making shortcuts.

I was using the product rule: [itex]\frac{d}{dX}[/itex](fg)=f[itex]^{|}[/itex]g+fg[itex]^{|}[/itex]
 
  • #9
okay so the first part doesn't vanish they both have x terms so it seems like you lost a a term :/ the natural log term
 
  • #10
Tenshou said:
okay so the first part doesn't vanish they both have x terms so it seems like you lost a a term :/ the natural log term

Confused :confused: can someone write out the correct solution?
 
  • #11
Is that the original problem?
 
  • #12
Sorry, we don't do that here at PF.

Look, you don't have a problem with optimization. You have a more basic problem with understanding how to take a derivative.
 
  • #13
SteamKing said:
Sorry, we don't do that here at PF.

Look, you don't have a problem with optimization. You have a more basic problem with understanding how to take a derivative.

I don't believe I have a problem with basic differentiation as I have studied it for well over a year and have been fine with it, however I am definitely not an expert. Where about do you believe the problem lies?
 
  • #14
I think I could have the answer if someone just helps me find x from: (x+ ln(1/2x))=0
 
  • #15
CallumC said:
(x+ ln(1/2x))=0

Is this the original question?
 
  • #16
Tenshou said:
Is this the original question?

Original Question:

A communications cable has a copper core with a concentric sheath of insulating material. If x is the ratio of the radius of the core to the thickness of the insulating sheath, the speed of a signal along the cable is given by:

S=8x2ln([itex]\frac{1}{2x}[/itex])

Find the value of x that gives the maximum speed.
 
  • #17
okay. ## S_{x}## is what you need that is the partial derivative of S with respect to x, so you get something that looks like this after taking the first derivative ##S_{x} = -16x ln( {2x} ) - 4x## So what I did was product rule then chain rule, I am not exactly sure if the last part with subtraction is right I am a little rusty with natural logs so that means that ##dS = S_{x} dx## that should be what you need. But there is still the problem of max. and what you can do to find the max is second derivative test... I think I am missing one term but I am not sure...
 
Last edited:
  • #18
Tenshou said:
okay. ## S_{x}## is what you need that is the partial derivative of S with respect to x, so you get something that looks like this after taking the first derivative ##S_{x} = -16x ln( {2x} ) - 4x## So what I did was product rule then chain rule, I am not exactly sure if the last part with subtraction is right I am a little rusty with natural logs so that means that ##dS = S_{x} dx## that should be what you need. But there is still the problem of max. and what you can do to find the max is second derivative test... I think I am missing one term but I am not sure...

Partial derivatives?? Why do you need them here. This is a single variable problem. Please don't confuse the OP by bringing up multivariable calculus.
 
  • #19
Tenshou: Don't help if you can't solve the problem.
 

What is optimisation?

Optimisation is the process of finding the best solution to a problem, usually by maximizing or minimizing a specific objective or goal.

Why is optimisation important?

Optimisation is important because it allows us to make the most efficient use of resources and achieve the best possible outcome. It is used in various fields such as engineering, economics, and computer science.

How do you approach an optimisation problem?

The first step in approaching an optimisation problem is to clearly define the objective or goal. Then, identify all the variables and constraints involved in the problem. Next, choose an appropriate mathematical model or algorithm to solve the problem. Finally, interpret and evaluate the results to determine the optimal solution.

What are some common techniques used in optimisation?

Some common techniques used in optimisation include linear programming, dynamic programming, genetic algorithms, and simulated annealing. These techniques use mathematical and computational methods to find the optimal solution to a problem.

What are the limitations of optimisation?

One limitation of optimisation is that it relies on assumptions and simplifications, which may not accurately reflect real-world situations. Additionally, it may be computationally intensive and time-consuming to find the optimal solution for complex problems. There is also a risk of overfitting the data and finding a solution that is not applicable in practical scenarios.

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