设a,b是实数,则a方+ab+b方-a-2b的最小值是

Question

设a,b是实数,则a方+ab+b方-a-2b的最小值是

Answer 1

设2元函数 f(a,b) = a^2 + ab + b^2 - a - 2b
令
f'_a = 2a + b - 1 = 0
f'_b = a + 2b - 2 = 0
得 a = 0, b = 1.
又,
f''_a_a = 2
f''_a_b = 1
f''_b_b = 2
f''_a_a * f''_b_b - (f''_a_b)^2 = 3 > 0.
所以,
2元函数 f(a,b) = a^2 + ab + b^2 - a - 2b
在a = 0, b = 1时达到最小值f(0,1)= -1.
如果不能用偏导数的知识.
可以配方.
a^2 + ab + b^2 - a - 2b
= (a + b/2)^2 + (3/4)b^2 - (a + b/2) - 3b/2
= (a + b/2 - 1/2)^2 - 1/4 + (3/4)[(b - 1)^2 - 1]
= (a + b/2 - 1/2)^2 + (3/4)[(b - 1)^2] - 1
>= -1
所以,最小值是 -1.

而且,在 a + b/2 = 1/2, b = 1时达到最小值.
也就是在 a = 0, b = 1时达到最小值-1.