Ralph White

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20 years, 307 days

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These are answers submitted by Ralph White

Sorry I mistaken the website is not this one

my apologies

nevetheless if any of you's gotspear time to give a look to the "image" and give me some suggestion about writing formulas for the posts I'll be glad and thankfull

Federiclet

HI

I gave an other try, solving by part the outcome is a fulll stop in front of arccos(x^2), not knowing any trigonometric identity wit linear argument i.e. Trig1(x^2) ->Trig2(n*x).

I include the URL or whatever it is for the image with the steps I manage to go through.

 

Here there should be the image I uploaded on my account (but is not here ... why?????)

NBNB please can anyone tell me how to get decent written formulas into posts?

I'll be gratefull someone could hel me about it.

cheers

F.

(too many informatic No, Niet , Nisba.. and is Sunday! ssssh.)

HI,

I'not a great Maple user,

still, I guess you should proceed integrating maple and classical min-max approach to functions. What I'm trying to say is that  working within the implicit frame, you'll nedd to know the diferential of your function (hope two dimensional), then ask Maple to optimize the result for minimal and maximal. Or symply equate to 0 the differential, I think after rewriting the funbction explicitly in dy/dx, you'll have couples of points (xi,yi) suitable for being min or max. The second derivative in the neighborogh of the differential zeros will tell you the change of convexity of the curve. ie for two zeros youll end up in sequence (y''left, y'=0, y''right ) with lets say  (+ 0 +, - 0 - ) or (+ 0 +, - 0 +): in the first case you have a max and a min, in the second case a max and a flex point. If you have more variables I think the jacobian of the function or the gradient of the function evaluated independently for each variabel (suppose all the other variable constant and derive for the one you choose as free, and so on, in turn) and solved for each of the variables asking each time the gradient be 0. Then, for say x1, x2, x3, or a sets of variable values (x1i),(x2i) ...(xji) by assembling the data togeter by hand you'll be able to see if there is any overlapping domain (in the second case), as a delimitation of the domain were the function has is  min and/or max. The data btained from the previous derivations will allow you to determine each min or max precisely of the curve).

Sorry  too many word (ITA)

And sorry non been able to be more precise.

peraphs if you could write down your equation I'll get somethiong more specific to think and work about, ...

feel free

take care

f

 

 

HI,

I'not a great Maple user,

still, I guess you should proceed integrating maple and classical min-max approach to functions. What I'm trying to say is that  working within the implicit frame, you'll nedd to know the diferential of your function (hope two dimensional), then ask Maple to optimize the result for minimal and maximal. Or symply equate to 0 the differential, I think after rewriting the funbction explicitly in dy/dx, you'll have couples of points (xi,yi) suitable for being min or max. The second derivative in the neighborogh of the differential zeros will tell you the change of convexity of the curve. ie for two zeros youll end up in sequence (y''left, y'=0, y''right ) with lets say  (+ 0 +, - 0 - ) or (+ 0 +, - 0 +): in the first case you have a max and a min, in the second case a max and a flex point. If you have more variables I think the jacobian of the function or the gradient of the function evaluated independently for each variabel (suppose all the other variable constant and derive for the one you choose as free, and so on, in turn) and solved for each of the variables asking each time the gradient be 0. Then, for say x1, x2, x3, or a sets of variable values (x1i),(x2i) ...(xji) by assembling the data togeter by hand you'll be able to see if there is any overlapping domain (in the second case), as a delimitation of the domain were the function has is  min and/or max. The data btained from the previous derivations will allow you to determine each min or max precisely of the curve).

Sorry  too many word (ITA)

And sorry non been able to be more precise.

peraphs if you could write down your equation I'll get somethiong more specific to think and work about, ...

feel free

take care

f

 

 

Could someone tell me why, if I enter a unit in 'pound'/'inch' Maple returns 'pound'/'ft'? What is the work around for this problem? Here is the problem cut and paste from my worksheet > sigma[p] := Units[Natural][`*`](85000, Unit(Units[Natural][`*`]('pound', Units[Natural][`/`](Units[Natural][`^`]('inch', 2))))); print(`output redirected...`); # input placeholder / 'lb' \ 12240000 Units:-Unit|-------| | 2| \('ft') /
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