己知x,y为正实数,且xy=4x+y+12,求xy最小值

显示全部楼层 · 2021-10-26 13:58:20

y=(4x+12)/(x-1)xy=x(4x+12)/(x-1)=4(x^2+3x)/(x-1)令t=x-1不等于0,现不妨设t>0x=t+1xy=4((t+1)^2+3(t+1))/t=4[(t^2+5t+4)/t]=4[5+(t+4/t)]t+4/t>=2根号(t*4/t)=4xy>=4[5+4]=36若t=2√(4xy)+12=4√(xy)+12,令t=√(xy)>0,有t^2-4t-12>=0,得t=6,所以xy的最小值为36

千问 · 2021-10-26 13:58:20

根据4x+y>=4√xy,故4x+y+12>=4√xy+12所以xy>=4√xy+12，移项得xy-4√xy-12>=0所以xy>=36

千问 · 2021-10-26 13:58:20

因为4x+y>=2√4xy=4√xy , 并且4x+y=12-xy所以12-xy>=4√xy 令√xy =t则12-t＾2>=4t解得-6<=t<=2即0<=√xy<=2所以0<=xy<=4即xy有最大值4.