设实数a,b满足:3a²-10ab+8b²+5a-10b=0,求n=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n=9a²+72b+2的最小值
答
式子化简(a-2b)(3a-4b)+5(a-2b)=0(a-2b)(3a-4b+5)=0当a-2b=0n=36b^2+72b+2=36(b+1)^2-34,最小值-34当3a-4b+5=0n=16b^2+32b+27=16(b+1)^2+11,最小值为11当都为0a=1,b=0.5n=47所以N最小值为-34