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Mathematica Help
Well, first hi!! This is my first post!! :D
Now, onto the real situation. I am currently in Calculus 4 in university and we have Mathematica projects.. problem is, I never learned how to really use it, and I've been flying on tutorials last semester, and the last project we did. But now, I have no tutorial, and I'm struggling.. badly.
So, the project is finding absolute minima and maxima [its from Neumann and Miller, project 5.8]
If I can just get started on it, that would be great.
the first part gives you a function and a constraining function. I have to find the critical points under the constraining function, then the minima and maxima on the unit sphere, and then on the closed ball. and its using lagrange multipliers.
I know this is kinda vague, I can def provide more information..
Anyway, more help would be AWESOME
Thanks guys
ETA: project file

Here's (a) in two parts. Had to convert to raw format to paste it here. I think you can follow it right? The D[f[x,y,z],x] is just the partial with respect to x. Same dif for others. The syntax:
extremaInUnitBall = Select[myExtrema,
Sqrt[#1[[1]]^2 + #1[[2]]^2 +
#1[[3]]^2] < 1 & ]
means "select from the table of myExtrema, those values for which
Code:
f[x_, y_, z_] := Sqrt[48]*x*y*z 
(1  x^2  y^2  z^2)^(3/2)
g[x_, y_, z_] := x^2 + y^2 + z^2  1
myExtrema = {x, y, z} /.
Solve[{D[f[x, y, z], x] == 0,
D[f[x, y, z], y] == 0,
D[f[x, y, z], z] == 0}, {x, y, z}]
extremaInUnitBall = Select[myExtrema,
Sqrt[#1[[1]]^2 + #1[[2]]^2 +
#1[[3]]^2] < 1 & ]
{{1, 0, 0}, {0, 1, 0}, {0, 0, 1},
{0, 0, 0}, {0, 0, 1}, {0, 1, 0},
{1, 0, 0}, {(Sqrt[3]/5), (Sqrt[3]/5),
(Sqrt[3]/5)}, {(Sqrt[3]/5),
Sqrt[3]/5, Sqrt[3]/5},
{Sqrt[3]/5, (Sqrt[3]/5), Sqrt[3]/5},
{Sqrt[3]/5, Sqrt[3]/5, (Sqrt[3]/5)}}
{{0, 0, 0}, {(Sqrt[3]/5), (Sqrt[3]/5),
(Sqrt[3]/5)}, {(Sqrt[3]/5),
Sqrt[3]/5, Sqrt[3]/5},
{Sqrt[3]/5, (Sqrt[3]/5), Sqrt[3]/5},
{Sqrt[3]/5, Sqrt[3]/5, (Sqrt[3]/5)}}
Second part to (a) using Lagrange multipliers. Again, the variable myLagrange is just the table of values for which the four Lagrange equations are zero. Then the syntax
(f[#1[[1]], #1[[2]], #1[[3]]] & ) /@
myLagrange
means, use the "pure function" (f[#1[[1]], #1[[2]], #1[[3]]]&) and go throgh the table myLagrange and substitute the first three elements in each term into the function f(x,y,z). From that list I get the max is 4/3 and the min is 4/3.
Code:
In[83]:=
myF[x_, y_, z_, \[Lambda]_] := f[x, y, z] +
\[Lambda]*g[x, y, z];
myLagrange = {x, y, z, \[Lambda]} /.
Solve[{D[myF[x, y, z, \[Lambda]], x] == 0,
D[myF[x, y, z, \[Lambda]], y] == 0,
D[myF[x, y, z, \[Lambda]], z] == 0,
D[myF[x, y, z, \[Lambda]], \[Lambda]] == 0},
{x, y, z, \[Lambda]}]
(f[#1[[1]], #1[[2]], #1[[3]]] & ) /@
myLagrange
Out[84]=
{{(1/Sqrt[3]), (1/Sqrt[3]), 1/Sqrt[3],
2}, {(1/Sqrt[3]), 1/Sqrt[3],
(1/Sqrt[3]), 2}, {1/Sqrt[3],
(1/Sqrt[3]), (1/Sqrt[3]), 2},
{1/Sqrt[3], 1/Sqrt[3], 1/Sqrt[3], 2},
{1, 0, 0, 0}, {0, 1, 0, 0},
{0, 0, 1, 0}, {0, 0, 1, 0},
{0, 1, 0, 0}, {1, 0, 0, 0},
{(1/Sqrt[3]), (1/Sqrt[3]),
(1/Sqrt[3]), 2}, {(1/Sqrt[3]),
1/Sqrt[3], 1/Sqrt[3], 2},
{1/Sqrt[3], (1/Sqrt[3]), 1/Sqrt[3],
2}, {1/Sqrt[3], 1/Sqrt[3],
(1/Sqrt[3]), 2}}
Out[85]=
{4/3, 4/3, 4/3, 4/3, 0, 0, 0, 0, 0, 0,
(4/3), (4/3), (4/3), (4/3)}