Question

y=(x-a)²+（x-b）²（a，b为常数） 求最大值或最小值

Answer 1

y
=(x-a)²+（x-b）²
= 2X² -(2A+2B)X + A² + B²
=2[ X² - (A+B)X + (A²+B²)/2]
=2{ X² - 2[(A+B)/2]X + (A²+B²)/2}
=2{ [X - (A+B)/2] ²+ (A²+B²)/2 - (A+B)²/4}

显然，当X = (A+B)/2时Y有最小值
Y
= 2 [ 0 ²+ (A²+B²)/2 - (A+B)²/4 ]
= [2A²+2B² - (A+B)² ]/2
= A²/2+B²/2-AB

Answer 2

y=(x-a)²+（x-b）²
= 2x² -(2a+2b)x + a² + b²
=2[ x² - (a+b)x] + a² +b²
=2[x²- (a+b)x + (a+b)²/4-(a+b)²/4]+ a² + b²
=2[x- (a+b)/2] ²-(a+b)²/2 + a² + b²
=2[x- (a+b)/2] ²+(a-b)²/2 .

当x- (a+b)/2=0时,即x=(a+b)/2时,
y有最小值为:(a-b)²/2