求(1+t)*(27t+3)/((1+3t)^2)的最大值 t〉=0
1个回答

设这个式子为y,

整理后得到:y=3(1+t)(9t+1)/[(1+3t)^2]

=3(9t^2+10t+1)/[(1+3t)^2]

=[3(9t^2+6t+1)+12t]/(9t^2+6t+1)

=3+12t/(9t^2+6t+1)

=3+12/(9t+6+1/t)

要得到y(max)只需求出分母9t+6+1/t的最小值,6为常数,则求出9t+1/t最小值

因为t>=0,所以(3t-1)^2>=0,9t^2+1>=6t,9t+1/t>=6,只有在t=1/3最小

因此,当t=1/3时,y(max)=4