问题标题:
函数最值若x^2+y^2=169,则函数f(x,y)=(24y-10x+338)^0.5+(24y+10x+338)^0.5的最大值是多少?
更新时间:2024-04-28 00:32:56
问题描述:
函数最值
若x^2+y^2=169,则函数f(x,y)=(24y-10x+338)^0.5+(24y+10x+338)^0.5的最大值是多少?
胡庚回答:
WOSIKAOYIXIA
答案:10根号(26)taiwanliaomingtianxieguocheng!!!
设a=169,则338=2a,t=f(x,y)=(24y+2a-10x)^0.5+(24y+2a+10x)^0.5----(1)
(1)式两边平方,t^2=2(24y+2a)+2[(24y+2a)^2-100x^2]^0.5
[(24y+2a)^2-100x^2]^0.5=[(24y)^2+96ay+4a^2-100(a-y2)]^0.5=[576y^2+96*169y+100y^2+4a(a-25)]^0.5=[676y^2+2*48*13*13y+4*169(169-25)]^0.5=[(26y)^2+2*26*2*13*12y+4*13^2*12^2]^0.5
=[(26y+2*12*13)^2]^0.5=|26y+12*26|=26|y+12|
所以t^2=2(24y+2a)+2[(24y+2a)^2-100x^2]^0.5=48y+4a+2*26|y+12|=48y+4*169+52|y+12|
=48y+26^2+52|y+12|
x^2+y^2=169===>y^2-13^2=x^2>=0===>-13