x^2 + 3xy + y^2 = 60 , where x and y are real numbers, what is the maxium possible value for xy?
The answer is 12, but I dun get how the teacher got that. Can any of u guys show me the process(calculation) of how u got that answer?
10分的數學問題
2007-10-08 10:30 am
回答 (1)
2007-10-08 5:22 pm
✔ 最佳答案
Consider(x - y)^2 >= 0 ..(*)
x^2 + y^2 - 2xy >=0
x^2 + y^2 >= 2xy
x^2 + 3xy + y^2 = 60
2xy + 3xy <= 60
xy <= 12
Note that the "=" in (*) hold when x = y
so, max of xy = 12
參考: My math knowledgr
收錄日期: 2021-04-13 19:30:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071008000051KK00398