Question

若实数a,b,c满足a*2+b*2+c*2=1,则3ab-3bc+2c*2的最大值为____.

Answer 1

先加上1得
3ab-3bc+2c²+1-1
=3ab-3bc+2c²+a²+b²+c²-1
=3ab-3bc+3c²+a²+b²-1
=(a+3/2b)²-3bc+3c²+(1-9/4)b²-1
=(a+3/2b)²+(3c²-5/4b²-3bc)-1
=(a+3/2b)²+3(c²-5/12b²-bc)-1
=(a+3/2b)²+3(c-1/2b)²-5/4b²-1/4b²-1
=(a+3/2b)²+3(c-1/2b)²-3/2b²-1
要求它的最大值。需要让b最小。c最大
令b=0,有a²+3c²-1令a=0,c=1得
=0+3-1=2
所以最大值为2