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標題: Max. and min. value [打印本頁]

作者: KT6491 時間: 2010-9-20 06:04 PM 標題: Max. and min. value

從同學口中得知以下題目，但小弟實在無能為力，束手無策，希望有高手可以幫手解答。
Question:
Find the maximum and minimum values of y.
[attach]29171[/attach]

P.S. post錯區tim，麻煩版主幫手，唔該!

[ 本帖最後由 KT6491 於 2010-9-20 06:19 PM 編輯 ]

作者: Simon 時間: 2010-9-20 07:20 PM

-1 ≦sinx ≦1
-2 ≦ sinx - 1 ≦ 0

-1 ≦cosx ≦1
1≦cosx + 2≦3

Max. value = 上max / 下min = 0 / 1 = 0
Min. value = 上min / 下max = -2/3

作者: notsang1 時間: 2010-9-20 07:37 PM

引用:

原帖由 傑Simon 於 2010-9-20 07:20 PM 發表 -1 ≦sinx ≦1-2 ≦ sinx - 1 ≦ 0-1 ≦cosx ≦11≦cosx + 2≦3Max. value = 上max / 下min = 0 / 1 = 0Min. value = 上min / 下max = -2/3

作者: Simon 時間: 2010-9-20 07:46 PM 標題: 回復 3# notsang1 的帖子

做咩?

作者: Simon 時間: 2010-9-20 07:46 PM

你想講答案係 -2 ?

作者: 桃子 時間: 2010-9-21 11:14 AM

引用:

原帖由 傑Simon 於 2010-9-20 07:20 PM 發表
-1 ≦sinx ≦1
-2 ≦ sinx - 1 ≦ 0

-1 ≦cosx ≦1
1≦cosx + 2≦3

Max. value = 上max / 下min = 0 / 1 = 0
Min. value = 上min / 下max = -2/3

If the question is y=s/t, where s,t are non-negative and independent and you know the ranges for s,t respectively, then your method will work.
However, in this case, s=sin x-1 and t=cos x+2. They are not independent (as (s+1)^2+(t-2)^2=1) and s≦0, you can't get max/min values this way.

[ 本帖最後由桃子於 2014-10-8 11:35 PM 編輯 ]

作者: 桃子 時間: 2010-9-21 11:16 AM

引用:

原帖由 KT6491 於 2010-9-20 06:04 PM 發表
從同學口中得知以下題目，但小弟實在無能為力，束手無策，希望有高手可以幫手解答。
Question:
Find the maximum and minimum values of y.
29171

P.S. post錯區tim，麻煩版主幫手，唔該!

The way I find most straight-forward is certainly by differentiation.

However, if you haven't learnt it yet, then there is probably a way to get the answer by playing around with trigonometric identities.

[ 本帖最後由桃子於 2010-9-21 11:23 AM 編輯 ]

作者: Simon 時間: 2010-9-21 04:01 PM 標題: 回復 6# 桃子的帖子

actually, i think the only problem for this question is to find the minimum value
For maximum value, this method should be ok, right?

作者: firoazen 時間: 2010-9-21 05:09 PM

dy/dx = (cos(x))/(cos(x)+2)+((sin(x)-1) sin(x))/(cos(x)+2)^2
dy/dx=0......

max{y} = 0 at x = pi/2

min{y} = -4/3 at x = 2 pi-2 tan^(-1)(3)

Using Calculus Approach

Or Another Way: Applying Method of T.T. (Be confident,u can do this

)

作者: 桃子 時間: 2010-9-21 09:31 PM

引用:

原帖由 傑Simon 於 2010-9-21 04:01 PM 發表
For maximum value, this method should be ok, right?

In general, "max. value = 上max / 下min" will not work if top and bottom are not independent.
Even if top and bottom are independent, you need to take care about the sign.

However, there is an alternative way to argue that the max. is 0 in this particular case.

The numerator is non-positive while the denominator is positive. Hence, we can conclude that the fraction is non-positive. Furthermore, we see that the value 0 can be attained (when sin x=1). Therefore, 0 is the maximum value.

作者: tangw 時間: 2010-9-26 05:50 PM

For max. value, the upper is greatest while the lower part is smallest.
For min. value, the upper is smallest while the lower part is greatest.
This is the principle.

作者: firoazen 時間: 2010-9-26 06:47 PM 標題: 回復 11# tangw 的帖子

好廢姐…唔好呃帖…
Btw, could you ask ur classmate how we canfind the ans?

作者: 桃子 時間: 2010-9-26 09:17 PM

引用:

原帖由 tangw 於 2010-9-26 05:50 PM 發表
For max. value, the upper is greatest while the lower part is smallest.
For min. value, the upper is smallest while the lower part is greatest.
This is the principle.

This is generally FALSE!!! (unless both numerator and denominator are positive)

作者: firoazen 時間: 2010-10-18 07:23 PM 標題: 回復 14# peterkong123 的帖子

What is meant by "refer to the text book"?
Do you mean the solution of this kinda question?
Do you know how to do this?

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