Re: Problem finding maximum

*To*: mathgroup at smc.vnet.net*Subject*: [mg127596] Re: Problem finding maximum*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 5 Aug 2012 15:00:54 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

> The correct answer is at x=0, where Abs[f[0,aa1]]=0.0540933 Hi. To get x=0 at your stated full precision (no decimal point) then maybe... v=.7481//Rationalize 7481/10000 f[x_,a_] := (a^3-6 x-a^2 (4+x)+a (2+12 x-4 x^2))/(8 a) Maximize[{Abs[f[x,v]],0<=x<=v},x] {43274639/800000000, {x->0} } %//N {0.0540933,{x->0.}} Which, as mentioned, is: NMaximize[{Abs[f[x,v]],0<=x<=v},x] {0.0540933,{x->0.}} I thought this was interesting for your problem... v=.; Manipulate[Plot[{Abs[f[x,v]],Abs[f[0,v]]},{x,0,v}, PlotRange->{0,1}, GridLines->False], {v,$MachineEpsilon,3,1/10} ] At small values of your variable, the max is the right side of the graph, or when x = v. At around 0.7, the max is at x=0. (close to your value of .7481) And at around 1.7, the right side (x=v) is the max. The middle section is not maximized. If you were doing a lot of calculations, you could reduce this to using IF statements. (Below, I kept only the 2 reals) v=. Solve[Abs[f[0,v]]==Abs[f[v,v]],v][[-2;;]] {{v->1/5 (6-Sqrt[6])}, {v->1/5 (6+Sqrt[6])}} %//N {{v->0.710102},{v->1.6899}} (* Above values appear to check from the graph *) Using these 2 values, you would have 2 different x values giving the same max value. Using 1 of the values, we do get the same values: {Abs[f[0,v]],Abs[f[v,v]]}/.v->1/5 (6-Sqrt[6]) //N {0.0420204, 0.0420204} If we pick a value, and use Maximize, only 1 x value is given. v=1/5 (6-Sqrt[6]); Maximize[{Abs[f[x,v]],0<=x<=v},x] //FullSimplify {1/50 (7-2 Sqrt[6]), {x->1/5 (6-Sqrt[6])} } But... we could get that same max value using x=0. Abs[f[0,v]]//FullSimplify 1/50 (7-2 Sqrt[6]) Just thought it interesting. :>) = = = = = = = = = = = = Dana DeLouis Mac & Math 8 = = = = = = = = = = = = On Aug 4, 9:56 pm, Cisco Lane <travl... at yahoo.com> wrote: > I cannot seem to find the corrrect maximum for the absolute value of the following function in 0<=x<=aa1 > > aa1 = .7481 > > f[x_, a_] = (a^3 - 6 x - a^2 (4 + x) + a (2 + 12 x - 4 x^2))/(8 a) > > For example, FindMaximum[{Abs[f[x, aa1]], 0 <= x <= aa1}, x] gives {0.0317268, {x -> 0.7481}} > > The correct answer is at x=0, where Abs[f[0,aa1]]=0.0540933 but I have tried every maximization function I can think of and nothing will give me the correct answer. Can anyone help?