设实数a,b满足:3a2-10ab+8b2+5a-10b=0,求u=9a2+72b+2的最小值.

问题描述:

设实数a,b满足:3a2-10ab+8b2+5a-10b=0,求u=9a2+72b+2的最小值.

由3a2-10ab+8b2+5a-10b=5(a-2b)+(a-2b)(3a-4b)=(a-2b)(3a-4b+5)=0,(6分)所以a-2b=0,或3a-4b+5=0.(8分)①当a-2b=0,即a=2b时,u=9a2+72b+2=36b2+72b+2=36(b+1)2-34,于是b=-1时,u的最小值为-34,...