Question

高中数学，求最大值

Answer 1

y1+y2=2t/(3t^2+4),
y1y2=36/(3t^2+4),
w=(y1+y2)^2-4y1y2
=4t^2/(3t^2+4)^2-144/(3t^2+4)
=(4t^2-432t^2-576)/(3t^2+4)^2
=-(428t^2+576)/(3t^2+4)^2,
设u=x^2>=0,则
w=-(428u+576)/(3u+4)^2,
w'=-[428/(3u+4)^2-6(428u+576)/(3u+4)^3]
=-[428(3u+4)-6(428u+576)]/(3u+4)^3
=[1284u+1744]/(3u+4)^3>0,
所以w是增函数，
当u--->+∞时w--->0,所以w有上确界0，
所以√w有上确界0，无最大值。

追问

有最大值啊

追答

y1+y2=24t/(3t^2+4),
y1y2=36/(3t^2+4),
w=(y1+y2)^2-4y1y2
=576t^2/(3t^2+4)^2-144/(3t^2+4)
=(576t^2-432t^2-576)/(3t^2+4)^2
=(144t^2-576)/(3t^2+4)^2,
设u=x^2>=0,则
w=-144(u-4)/(3u+4)^2,
w'=144[1/(3u+4)^2-6(u-4)/(3u+4)^3]
=144[3u+4-6u+24)]/(3u+4)^3
=144[-3u+28]/(3u+4)^3，
所以u=28/3时，w取最大值3/4，
于是√w的最大值是√3/2.

追问

好的，谢谢

追答

别客气！