柯西不等式 高二二下第四冊?
回答 (1)
2021-03-16 8:12 pm
2x-3y+6z = 2(x-2)-3(y+1)+6z +7
[2(x-2)-3(y+1)+6z]^2
≦ [2^2+(-3)^2+6^2][(x-2)^2+(y+1)^2+z^2]
= 49^2
不等式之等號成立的充要條件是
(x-2)/2 = (y+1)/(-3) = z/6
由 (x-2)^2+(y+1)^2+z^2 = 49, 故
x-2 = 2t, y+1 = -3t, z = 6t,
而 t = ±1, 分別對應到 2(x-2)-3(y+1)+6z 的
最大值 49 與最小值 -49.
故 2x-3y+6z 之最大值為 49+7 = 56,
發生於 x = 4, y = -4, z = 6.
[2(x-2)-3(y+1)+6z]^2
≦ [2^2+(-3)^2+6^2][(x-2)^2+(y+1)^2+z^2]
= 49^2
不等式之等號成立的充要條件是
(x-2)/2 = (y+1)/(-3) = z/6
由 (x-2)^2+(y+1)^2+z^2 = 49, 故
x-2 = 2t, y+1 = -3t, z = 6t,
而 t = ±1, 分別對應到 2(x-2)-3(y+1)+6z 的
最大值 49 與最小值 -49.
故 2x-3y+6z 之最大值為 49+7 = 56,
發生於 x = 4, y = -4, z = 6.
收錄日期: 2021-05-04 02:36:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210315143529AAg5GeO