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Re: Local extrema of a function of two variables

  • To: mathgroup at smc.vnet.net
  • Subject: [mg90017] Re: Local extrema of a function of two variables
  • From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
  • Date: Thu, 26 Jun 2008 04:44:49 -0400 (EDT)

On Wed, Jun 25, 2008 at 4:56 PM, Barun von Doppeldecker
<nikoi8888 at gmail.com> wrote:

> if you still want to help. here is another problem i came across today: i
> have a function f(x, y) = 4xy â?? x^4 â?? y^4 and i have to do three things.
> First i have to find candidates for local extremes, then i have to exactly
> pinpoint that extremes and finally plot a graf surrounding that point of
> extremum.
> I have found candidates, but the other two things i just can't....Here is
> what i've done:
>
> f[x_,y_]:=4 x y-x^4-y^4
>
> Solve[{D[f[x,y],x]Å 0,D[f[x,y],y]Å 0},{x,y}]
>
> if you could help me with other 2 things, i would be really grateful... if
> not, thank you anyway.... ;)

<snip>

[... Reply cross-posted to MathGroup ...]

You could follow the following "template".

(* Say that the function we want to study is the following: *)

f[x_, y_] := 4 x y - x^4 - y^4

(* We compute the first derivatives of f w.r.t.to x and y: *)

gradient = D[f[x, y], {{x, y}, 1}]

{-4 x^3 + 4 y, 4 x - 4 y^3}

(* We equate each derivative to zero and solve for x and y: *)

sols = Solve[gradient == 0]

{{x -> -1, y -> -1}, {x -> 0, y -> 0}, {x -> -I,
  y -> I}, {x -> I, y -> -I}, {x -> 1, y -> 1}, {x -> -(-1)^(1/4),
  y -> -(-1)^(3/4)}, {x -> (-1)^(1/4),
  y -> (-1)^(3/4)}, {x -> -(-1)^(3/4),
  y -> -(-1)^(1/4)}, {x -> (-1)^(3/4), y -> (-1)^(1/4)}}

(* We want only the real points,so we discard the complex \
solutions: *)

pts =
 Cases[sols, {x -> vx_, y -> vy_} /; Im[vx] == 0 && Im[vy] == 0]

{{x -> -1, y -> -1}, {x -> 0, y -> 0}, {x -> 1, y -> 1}}

(* So we have the critical points.
Now,we compute the Hessian matrix, *)

hessian = D[f[x, y], {{x, y}, 2}]

{{-12 x^2, 4}, {4, -12 y^2}}

(* and take its determinant: *)

d = hessian // Det

-16 + 144 x^2 y^2

(* Then we compute the values of the Hessian determinant at \
the critical points: *)

d /. pts

{128, -16, 128}

D[f[x, y], {x, 2}] /. pts

{-12, 0, -12}

(* We apply the second derivative test.
Since the Hessian is negative at the point (0,0),we \
conclude that this is a saddle point.
Since the Hessian is positive at \
the points (-1, -1) ant (1,1),and the second derivatives w.r.t. x at
these points are negative, we conclude that those points are local maxima. *)

(* We can "check" (or anticipate) these results by \
plotting the function: *)

Plot3D[f[x, y], {x, -2, 2}, {y, -2, 2}, PlotRange -> All]

ContourPlot[f[x, y], {x, -2, 2}, {y, -2, 2},
 ColorFunction -> "SunsetColors"]

[keywords: function two variables local minimum maximum critical point
Hessian calculus]

Regards,
-- 
Jean-Marc


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