Find x value for max(y)

Forum|Forum|13 years ago
February 14, 2013
11 replies
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Hi, I have a function y[x which behaves as an inverted parabola. By using the command max(y) I can find the maximum y value, but I would also like to find the associated x value. What command can I use to find the x value? Thanks!

Other

Best answer by Werner_E

J M schrieb:
Thank you, In my case however the 'x' term is not in matrix form, it is a series. Please find the attached file to see the problem. I know that the maximum point can be found by setting the derivative of the function equal to zero but this will be impossible as the model grows in complexity.

As you do not want to turn your vector E into a function and use the derivative you have at least two options:

1) turn x into a function and use the method I had posted

2) use "match" to get the index if the maximal value which in your case would be the searched for x value, too.

See attached worksheet for details.

Edited: haven't realized that Stuart already had posted.

I second his remark about not giving up on the function approach. At the end of my worksheet you find a symbolic solution using functions. If the functions get more complex and the symbolic solver is not able to find a solution you would resort to a numeric solution by using a solve block.

have just updated the file and included a solve block and the root function

H

HarveyHensley

12-Amethyst

There may be a lot of ways to do this. One is to set f(x)=max_y in a Given - Find(x) block. Use the Boolean =. The max_y may need to be obtained from an earlier operation.

Werner_E

25-Diamond I

I think what you a looking for is the "lookup" function in Mathcad.

For future questions keep in mind that you will get a better response from the community if you attach a worksheet (upper right corner, extended editor). It also makes sense stating which version of MC you use.

From what you wrote I assume that you have two vectors, x and y.

What you need is something like x_max:=lookup( max(y), y, x )[0

Another solution would use the "match" command y[match(max(y),x)[0

Remember that match returns the index - in the screenshot below ORIGIN was 0, so index 2 is the third position.

lookup and match return vectors as there may be more than one occurence (as in the example below). Thats the reason for the "]0" in the two expressions above, assuming you want the first occurence of the max (and ORIGIN is left at 0),

P

ptc-4921095

Author

1-Visitor

Thank you, In my case however the 'x' term is not in matrix form, it is a series. Please find the attached file to see the problem. I know that the maximum point can be found by setting the derivative of the function equal to zero but this will be impossible as the model grows in complexity.

1_MAX-xmcd.zip

S

StuartBruff

23-Emerald V

J M wrote:
Thank you, In my case however the 'x' term is not in matrix form, it is a series. Please find the attached file to see the problem. I know that the maximum point can be found by setting the derivative of the function equal to zero but this will be impossible as the model grows in complexity.

You simply need to convert x into a vector (or create another variable based on x), eg

xx[x:=x

You can then use Werner's suggested method

lookup(E.max,E,xx)

However, you will face a slight problem here as E is sufficiently flat near the top of your curve that several values of E meet the default criterion of E.max. You will have to reduce TOL (eg, 10^-5) to get an answer.

Don't write off the derivative method just yet. You many find either the symbolic or numeric solve blocks can give you correct value ... the lookup method is constrained by the limited set of E values that can be generated as result of the step size of x

Stuart