A^b=64 min(a*b)=?

Interpretation 1: a > 0, b > 0 real numbers - Constraint: a^b = 64, so b = ln(64) / ln(a). - Minimize P = a b = a · [ln(64) / ln(a)] = (ln 64) · a / ln(a). - Let f(a) = a / ln(a). Then f'(a) = (ln(a) − 1) / (ln(a))^2, so f'(a) = 0 at ln(a) = 1 → a = e. - This gives b = ln(64) / ln(e) = ln(64). - Minimum value: P_min = a b = e · ln(64) = e · 6 ln 2 ≈ 11.305. So, the minimum is e · ln(64) attained at a = e, b = ln(64). Interpretation 2: a and b are positive integers - Representations of 64 as a^b with integers: (a,b) ∈ {(64,1), (8,2), (4,3), (2,6)}. - Products: 64, 16, 12, 12. The minimum is 12, achieved by (a,b) = (4,3) or (2,6). Remark on allowing negative a - If b is even and a = −64^{1/b}, a^b = 64, and a b can go to −∞ as b grows, so there is no minimum in that (loose) case.