【微積分】極值
2014-03-15 12:14 am
Find the extreme value for f(x,y) = 4 + xy - (x^2) - (y^2) on S = {(x,y) : (x^2) + (y^2) ≤ 1}.
回答 (2)
2014-03-15 4:12 am
✔ 最佳答案
請參考下列作法：.................圖片參考：https://s.yimg.com/rk/AD07982684/o/1982597079.png
2014-03-14 20:16:55 補充：
倒數第5列筆誤修正
D = 1 - 4 = -3 < 0
2014-03-15 3:17 am
先取等號來做,然後再來檢驗是否<=1由限制條件: y^2=1-x^2 => y=+-√(1-x^2); x<=1取y=+√(1-x^2) {y=-√(1-x^2)類推}g(x)=4+x√(1-x^2)-x^2-(1-x^2)=3+x√(1-x^2)g'(x)=√(1-x^2)-x^2/√(1-x^2)=(1-x^2-x^2)/√(1-x^2)=(1-2x^2)/√(1-x^2)=0x^2=1/2 => x=+-1/√2=> y=+-√(1-x^2)=+-1/√2
檢驗合格: x^2+y^2=1/2+1/2<=1 OK
max=f(1/√2,1/√2)=f(-1/√2,-1/√2)
=4+1/2-1/2-1/2=7/2.....ans
min=f(1/√2,-1/√2)=f(-1/√2,1/√2)
=4-1/2-1/2-1/2=5/2.....ans
檢驗合格: x^2+y^2=1/2+1/2<=1 OK
max=f(1/√2,1/√2)=f(-1/√2,-1/√2)
=4+1/2-1/2-1/2=7/2.....ans
min=f(1/√2,-1/√2)=f(-1/√2,1/√2)
=4-1/2-1/2-1/2=5/2.....ans
收錄日期: 2021-05-02 11:11:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140314000016KK03917