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I need to maximize the following function. Unfortunately I have to maximize without using numerical methods to prove that the maximal value is 3Sqrt(2Sqrt(3)-3) (about 2.04375). I tried it twice (once with a computer with 8 gm ram (i5 proc.), and with a computer with 16 gm of ram (i7)), but in both cases the memory got full. Is it possible that this can be computed? If so what machine do I need?

v[q_, w_, e_, r_, t_, z_, u_, i_, o_, p_, a_, s_, d_, f_, g_, h_] := 
1/12 (Sqrt[2] Cos[o] Cos[s] Sin[p] + 
   2 Cos[i] Cos[o] Sin[g] Sin[h] Sin[p] - 
   2 Cos[o] Cos[s] Sin[g] Sin[h] Sin[p] + 
   Sqrt[2] Cos[i] Sin[o] Sin[p] - Sqrt[2] Cos[s] Sin[o] Sin[p] + 
   Sqrt[2] (Cos[u] (Cos[p] - Cos[w]) Sin[i] - Cos[i] Cos[o] Sin[p]) -
    Sqrt[2] Cos[e] Cos[s] Sin[r] + Sqrt[2] Cos[e] Cos[w] Sin[r] + 
   Sqrt[2] Cos[s] Sin[e] Sin[r] - Sqrt[2] Cos[w] Sin[e] Sin[r] + 
   2 Cos[e] Cos[s] Sin[d] Sin[f] Sin[r] - 
   2 Cos[e] Cos[z] Sin[d] Sin[f] Sin[r] - 
   Sqrt[2] Cos[a] Cos[p] Sin[s] + Sqrt[2] Cos[a] Cos[r] Sin[s] + 
   Sqrt[2] Cos[p] Sin[a] Sin[s] - Sqrt[2] Cos[r] Sin[a] Sin[s] + 
   2 Cos[a] Cos[h] Sin[d] Sin[f] Sin[s] - 
   2 Cos[a] Cos[r] Sin[d] Sin[f] Sin[s] - 
   2 Cos[a] Cos[f] Sin[g] Sin[h] Sin[s] + 
   2 Cos[a] Cos[p] Sin[g] Sin[h] Sin[s] + 
   2 Cos[h] Cos[o] Sin[a] Sin[p] Sin[s] - 
   2 Cos[a] Cos[h] Sin[o] Sin[p] Sin[s] - 
   2 Cos[e] Cos[f] Sin[a] Sin[r] Sin[s] + 
   2 Cos[a] Cos[f] Sin[e] Sin[r] Sin[s] - 
   2 Cos[g] Sin[
     h] (-Cos[z] Sin[d] Sin[f] + Cos[i] Sin[o] Sin[p] + 
      Cos[s] (Sin[d] Sin[f] - Sin[o] Sin[p]) - Cos[f] Sin[a] Sin[s] +
       Cos[p] Sin[a] Sin[s] + (-Cos[p] + Cos[z]) Sin[i] Sin[u]) + 
   Sqrt[2] Cos[i] Cos[q] Sin[w] - Sqrt[2] Cos[q] Cos[r] Sin[w] - 
   Sqrt[2] Cos[i] Sin[q] Sin[w] + Sqrt[2] Cos[r] Sin[q] Sin[w] - 
   2 Cos[q] Cos[z] Sin[e] Sin[r] Sin[w] + 
   2 Cos[e] Cos[z] Sin[q] Sin[r] Sin[w] - 
   2 Cos[h] Cos[t] Sin[d] Sin[f] Sin[z] + 
   2 Cos[r] Cos[t] Sin[d] Sin[f] Sin[z] - 
   2 Cos[f] Cos[t] Sin[e] Sin[r] Sin[z] + 
   2 Cos[t] Cos[w] Sin[e] Sin[r] Sin[z] + 
   2 Cos[f] Sin[h] Sin[g - t] Sin[z] - 
   2 Cos[i] Sin[h] Sin[g - t] Sin[z] + 
   2 Cos[e] Cos[f] Sin[r] Sin[t] Sin[z] - 
   2 Cos[e] Cos[w] Sin[r] Sin[t] Sin[z] + 
   2 Cos[i] Cos[t] Sin[q] Sin[w] Sin[z] - 
   2 Cos[r] Cos[t] Sin[q] Sin[w] Sin[z] - 
   2 Cos[i] Cos[q] Sin[t] Sin[w] Sin[z] + 
   2 Cos[q] Cos[r] Sin[t] Sin[w] Sin[z] + 
   Sin[i] (2 Cos[h] Sin[p] Sin[o - u] + Sqrt[2] Cos[w] Sin[u] - 
      Cos[p] (2 Cos[u] Sin[g] Sin[h] + Sqrt[2] Sin[u]) + 
      2 Cos[z] (Cos[u] Sin[g] Sin[h] - Sin[q - u] Sin[w]) + 
      2 (-Cos[h] + Cos[w]) Sin[t - u] Sin[z]) + 
   2 Cos[d] Sin[
     f] ((Cos[s] - Cos[z]) (Sin[g] Sin[h] - 
         Sin[e] Sin[r]) - (Cos[h] - Cos[r]) (Sin[a] Sin[s] - 
         Sin[t] Sin[z])))
X = Maximize[{v[q, w, e, r, t, z, u, i, o, p, a, s, d, f, g, h], 
  0 <= q <= 2 Pi, 0 <= e <= 2 Pi, 0 <= t <= 2 Pi, 0 <= u <= 2 Pi, 
  0 <= o <= 2 Pi, 0 <= a <= 2 Pi, 0 <= d <= 2 Pi, 0 <= g <= 2 Pi, 
  0 <= w <= Pi, 0 <= r <= Pi, 0 <= z <= Pi, 0 <= i <= Pi, 
  0 <= p <= Pi, 0 <= s <= Pi, 0 <= f <= Pi, 0 <= h <= Pi}, {q, w, e, 
  r, t, z, u, i, o, p, a, s, d, f, g, h}]
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  • $\begingroup$ Try NMaximize $\endgroup$ – Ulrich Neumann Nov 5 '20 at 15:44
  • $\begingroup$ I see the following way. First, work with an expression instead of a function: v=1/12 (Sqrt[2] Cos[o] Cos[s] Sin[p] + .... Second, apply TrigExpand@Expand[v]. Third, make a change {Cos[g]->cg,Sin[g]->sg,...} and add the restrictions cg^2+sg^2==1&&...&&sf>=0&& (sf>=0 replaces 0 <= f <= Pi). Therefore, you reduce the transcendental problem under consideration to the polynomial problem with 32 variables. This is still too much for my comp, but yours is more powerful. Good luck! $\endgroup$ – user64494 Nov 5 '20 at 20:35
  • $\begingroup$ This is a great idea! Thanks! $\endgroup$ – csaba.medgyes Nov 5 '20 at 22:04
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Try NMaximize

X = NMaximize[{v[q, w, e, r, t, z, u, i, o, p, a, s, d, f, g, h], 
0 <= q <= 2 Pi, 0 <= e <= 2 Pi, 0 <= t <= 2 Pi, 0 <= u <= 2 Pi, 
0 <= o <= 2 Pi, 0 <= a <= 2 Pi, 0 <= d <= 2 Pi, 0 <= g <= 2 Pi, 
0 <= w <= Pi, 0 <= r <= Pi, 0 <= z <= Pi, 0 <= i <= Pi, 
0 <= p <= Pi, 0 <= s <= Pi, 0 <= f <= Pi, 0 <= h <= Pi}, {q, w, e, 
r, t, z, u, i, o, p, a, s, d, f, g, h}]

(*{2.04375, {q -> 1.93316, w -> 1.10085, e -> 0.0556802, r -> 0.266313, 
t -> 3.24438, z -> 1.2329, u -> 2.2633, i -> 2.25653, o -> 0.121048,
p -> 2.65805, a -> 5.69525, s -> 1.58568, d -> 4.55036, 
f -> 1.12351, g -> 4.20617, h -> 2.27592}}*)
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  • $\begingroup$ Thanks for the fast response, I tried and it worked, but unfortunately they do not accept numerical solutions $\endgroup$ – csaba.medgyes Nov 5 '20 at 15:49
  • $\begingroup$ Sorry, didn't read your question comletely... $\endgroup$ – Ulrich Neumann Nov 5 '20 at 16:21

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