Question

求解求过程 高二数学

ab为实数

Answer 1

F = a^2+b^2+k(a^2-ab-2b^2-1)
F'<a> = 0 : 2a+2ka-b = 0 (1)
F'<b> = 0 : 2b-a-4kb = 0 (2)
F'<k> = 0 : a^2-ab-2b^2 = 1 (3)
由(1) , b = 2a(1+k) 代入 (2) 得
4a(1+k)-a-8ka(1+k) = 0, 由 (3) , a ≠ 0;
得 8k^2+4k-3 = 0, 即 k = (-1±√7)/4.
k = (-1+√7)/4 时
b = 2a(1+k) = a(3+√7)/2, 代入(3) 得
a^2-a^2 (3+√7)/2 - 2a^2[(3+)/2]^2 = 1， 无解。
k = (-1-√7)/4 时
b = 2a(1+k) = a(3-√7)/2, 代入(3) 得
a^2-a^2 (3-√7)/2 - 2a^2[(3-√7)/2]^2 = 1
解得 a^2 = (10+7√7)/27，
b^2 = [(3-√7)/2]^2 a^2 = [(8-3√7)/2](10+7√7)/27 = (26√7-67)/54
最小 a^2+b^2 = (40√7-47)/54