求(t+1)(9t+1)/(3t+1)^2最大值(t>0)

高中学生可以这样做：
对(t+1)(9t+1)/(3t+1)^2求导,得(18t+10)(3t+1)的平方-(18t+6)(9t的平方+10t+1)/(3t+1)的4次方
化简得：（-12t+4）/（3t+1)的3次方
当：（-12t+4）/（3t+1)＝0时,t＝1/3
t0
t>1/3时,（-12t+4）/（3t+1)

y=(t+1)(9t+1)/(3t+1)^2
=(9t^2+10t+1)/(9t^2+6t+1)
=[(9t^2+6t+1)+4/3*(3t+1)-4/3]/(3t+1)^2
=1+4/3*1/(3t+1)-4/3*1/(3t+1)^2
设x=1/(3t+1),t>0,则有0y=1+4/3x-4/3x^2=-4/3(x-1/2)^2+4/3*1/4+1=-4/3*(x-1/2)^2+4/3
即当x=1/2时，Y有最大值是：4/3
即当t=1/3时，Y有最大值是：4/3