Mathematica Asked by csaba.medgyes on July 2, 2021
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}]
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}}*)
Answered by Ulrich Neumann on July 2, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP