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Use a numerical approach: NMinimize[{1/2 x^2 y^2 + y^2 z^2 + z^2 x^2 + 96/(x + y + z + 1), x >= 0 && y >= 0 && z >= 0 && x^2 + y^2 + z^2 == 5}, {x, y, z}] {24.6693, {x -> 2.00209, y -> 0 …

Your code works here (Mathematica v 9): x = Table[Symbol[StringJoin["x", ToString[i]]], {i, 7}]; A = {3, 2, 5, 1, 7, 9, 6}; Minimize[{A.x, (Apply[And, Thread[0 <= x <= 1]]) && (Apply[Plus, x] ==3) && …

Maximize[{Log[n]^(1/Log[n]), n > 1}, n] (* {E^(1/E), {n -> E^E}} *)

Following the trend of posting alternative methods and skipping the obvious NMaximize[{t, sol[x, t] == 1, -58 <= x <= 50, 50 <= t <= 58}, {x, t}] max = SortBy[PixelValuePositions[ i = Binariz …

Only for this question not getting to the un-answered queue: Show[ContourPlot3D[ios, {ac, 0, 2}, {ah, 0, 2}, {ap, 0, 2}, Contours -> 10], Graphics3D@({Red, PointSize[0.01], Line[pts …

w = RandomReal[{0, 1}, 10]; b = RandomReal[{0.5, 1}, 10]; a = RandomReal[{0, .5}]; rr = Array[r, 10]; Minimize[{w.Log@rr, Thread[a <= rr <= b]}, rr]

FullSimplify[ ArgMax[{a*Sqrt[x] + b*Sqrt[y], x + y <= 10 && a > 0 && b > 0}, {x, y}], a > 0 && b > 0] (* {(10 a^2)/(a^2 + b^2), (10 b^2)/(a^2 + b^2)} *)

Your formulation doesn't work because your constraints don't involve variables. The preferred solution was posted as a comment to your question, but for making your original one work you could do: NM …

As Szabolcs stated, Maximize is calling NMaximize. The problem is that the call is not being done with appropriate options for your case. Just compare for example: NMaximize[{x*(1 - 0.01 x), x ∈ Int …

I am not sure if I am following your problem. Perhaps this is a partial answer. For solving ODEs like yours for a variable number of functions, you could do something like: dims = 3; k = RandomInteg …

Using FindAllCrossings from here norm[x_List] := Sqrt[x.x] (* Norm[ ] is always problematic *) r[t_] := {3 Cos[3.3 π t], Sin[4 π t] + 4 t}; n[t_] := norm[r'[t] r''[t]]/norm[r'[t]]; (* now we find th …

If you want to use the results from Reduce, you could do: Max[j /. Solve@ Reduce[2^j/(j + 1) <= 10, j, Integers]]

For example: experimArray = Range@20; theoretArray = ListConvolve[{a, b, c, d, e}, Range@24] sol = NMinimize[Norm[experimArray - theoretArray], {a, b, c, d, e}] So: theoretArray /. sol[[2]] (* {1. …

You may use FindFit l = {0.344027 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-1.3439 - 0.0174533 c)], 0.679511 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (0.20944 - 0.0174533 c)], 0.436543 == 0.5 (a + b) …

Diophantine problems are tough and there is no silver bullet. In your example this works: i = IntegerPart; sol = NMaximize[{(3 i@ n + 4)/(2 i@n + 1), n > 1}, n]; i@n /. sol[[2]] (* 1 *)