[THOUGHT] I will use Python to compute the maximum value of the function F over the vertices of the set S, then express that maximum as a fraction p/q in lowest terms and compute p^2 * q. [/THOUGHT]

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[THOUGHT] I will enumerate all the vertices of the polyhedron defined by the constraints and evaluate the function F at each vertex to find the maximum value M. Then, I will express M as a fraction p/q in lowest terms and compute p² * q. [/THOUGHT]

[PYTHON] from itertools import combinations import numpy as np from fractions import Fraction

Define the planes: each plane is a tuple (a, b, c, d) for the equation ax + by + cz = d

planes = [ (1, 0, 0, 0), # x = 0 (0, 1, 0, 0), # y = 0 (0, 0, 1, 0), # z = 0 (1, 1, 0, 2), # x + y = 2 (1, 2, 0, 3), # x + 2y = 3 (1, 0, 1, 3), # x + z = 3 (0, 1, 1, 3) # y + z = 3 ]

Function F

def F(x, y, z): return -5x**2 - 4y2 - 6*z2 + 4xy - 2xz + 3yz + 20x + 14y + 15*z + 1

Check if a point (x, y, z) is in S

def in_S(x, y, z): if x < 0 or y < 0 or z < 0: return False if x + y > 2: return False if x + 2*y > 3: return False if x + z > 3: return False if y + z > 3: return False return True

Enumerate vertices by solving systems of 3 planes

vertices = [] for combo in combinations(range(7), 3): A = np.array([planes[i][:3] for i in combo]) b = np.array([planes[i][3] for i in combo]) try: sol = np.linalg.solve(A, b) x, y, z = sol if in_S(x, y, z): vertices.append((x, y, z)) except np.linalg.LinAlgError: continue

Evaluate F at each vertex and find maximum

max_val = -float('inf') for pt in vertices: val = F(pt[0], pt[1], pt[2]) if val > max_val: max_val = val

Convert max_val to fraction

frac = Fraction(max_val).limit_denominator() p = frac.numerator q = frac.denominator

Compute p^2 * q

result = p**2 * q print(result) [/PYTHON]

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