x,y∈(0,+∞),x+2y+xy=30,求x+y的取值范围答案是x+y∈[8√2-3,30),
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/26 05:03:15
![x,y∈(0,+∞),x+2y+xy=30,求x+y的取值范围答案是x+y∈[8√2-3,30),](/uploads/image/z/5246122-58-2.jpg?t=x%2Cy%E2%88%88%280%2C%2B%E2%88%9E%29%2Cx%2B2y%2Bxy%3D30%2C%E6%B1%82x%2By%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E7%AD%94%E6%A1%88%E6%98%AFx%2By%E2%88%88%5B8%E2%88%9A2-3%2C30%29%2C)
x,y∈(0,+∞),x+2y+xy=30,求x+y的取值范围答案是x+y∈[8√2-3,30),
x,y∈(0,+∞),x+2y+xy=30,求x+y的取值范围
答案是x+y∈[8√2-3,30),
x,y∈(0,+∞),x+2y+xy=30,求x+y的取值范围答案是x+y∈[8√2-3,30),
x+2y+xy=30,
则y=(30-x)/(x+2),
因为y>0, (30-x)/(x+2) > 0,
所以0<x<30.
设x+2=t,则x=t-2,2<t<32.
x+y= x+(30-x)/(x+2)
=t-2+(32-t)/t
=t-2+32/t-1
= t+32/t-3……利用基本不等式
≥2√(t•32/t)-3=8√2-3.
函数t+32/t在[0,4√2]上递减,在[ 4√2,+∞)上递增,
因为2<t<32,所以t=32时,t+32/t最大,t+32/t-3<30.
∴x+y∈[8√2-3,30).
(x+2)(y+1)=32 则,x+2+y+1变成x+2+32/(x+2),考虑函数单调性可得最小值8√2,最大值32~然后就是结果了~