Max. & Min

我不是已經答了你嗎？

在 https://hk.knowledge.yahoo.com/question/question?qid=7015072400046

y can even achieve +∞ if x can be any real number.

You can consider setting x tend to a value such that tan x approaches -1/2

2015-07-27 16:30:47 補充：
I see, you did not state clearly it is asking for LOCAL maximum.

Yes, there are local extremums, please see:

https://www.wolframalpha.com/input/?i=%2818+sin+x+cos+x%29%2F%282+sin+x+%2B+cos+x%29

2015-07-29 02:32:52 補充：
邊位都好 老師：
18 ∛2 / (4 + ∛2 + 2∛4) ≈ 2.68871645

However, when x = 0.670888,
y = 18 sin x/(1 + 2 tan x) ≈ 4.32490806 > 2.68871645

according to the page in Comment 004 and my own calculation.

2015-07-29 02:53:13 補充：
邊位都好 老師：
我知道問題所在了，我重計時也跟你一樣犯了相同的錯誤。

Solve to tan x = 1/∛2.

Then draw the right-angled triangle.

The hypotenuse is NOT 1 + ∛4, but actually √(1 + ∛4).

We both forgot the square root of the Pythagoras theorem.

2015-07-29 02:57:01 補充：
The local maximum of y is
[18/√(1 + ∛4)] / (1 + 2/∛2) ≈ 4.32490806.

This is attained when x = tan⁻¹(1/∛2) ≈ 0.670888.

That is, when tan x = 1/∛2 and sin x = 1/√(1 + ∛4).

2015-07-29 03:06:26 補充：
★ The answer is given in Comment 008 ★

Below is the calculation of the differentiation part:

a = 18 sin x/(1 + 2 tan x)

da/dx
= 18 d/dx [sin x/(1 + 2 tan x)]
= 18/(1 + 2 tan x)² [(1 + 2 tan x)cos x - sin x(2 sec² x)]
= 18/(1 + 2 tan x)² (cos x + 2 sin x - 2 sin x/cos² x)]

2015-07-29 03:10:07 補充：
= 18/[(1 + 2 tan x)² cos²x] (cos³x + 2 sin x cos²x - 2 sin x)
= 18/[(1 + 2 tan x)² cos²x] [cos³x - 2 sin x (1 - cos²x)]
= 18/[(1 + 2 tan x)² cos²x] (cos³x - 2 sin³x)
= 18/[(1 + 2 tan x)² cos²x] (cos x - ∛2 sin x) (cos²x + ∛2 cos x sin x + ∛4 sin²x)

da/dx = 0 (for max) ⇒ cos x - ∛2 sin x = 0

2015-08-01 02:49:53 補充：
Please also read the comment and the last post.

Answer:
The local maximum of y is
[18/√(1 + ∛4)] / (1 + 2/∛2) ≈ 4.32490806.

This is attained when x = 2nπ + tan⁻¹(1/∛2) for any integer n
where tan⁻¹(1/∛2) ≈ 0.670888.
That is, when tan x = 1/∛2 and sin x = 1/√(1 + ∛4).

Note:
Since the pattern of 18 sin x/(1 + 2 tan x) is periodic, there are infinitely many values of x that make the function attains the same local maximum.

Details:
y = 18 sin x/(1 + 2 tan x)

dy/dx
= 18 d/dx [sin x/(1 + 2 tan x)]
= 18/(1 + 2 tan x)² [(1 + 2 tan x)cos x - sin x(2 sec² x)]
= 18/(1 + 2 tan x)² (cos x + 2 sin x - 2 sin x/cos² x)]
= 18/[(1 + 2 tan x)² cos²x] (cos³x + 2 sin x cos²x - 2 sin x)
= 18/[(1 + 2 tan x)² cos²x] [cos³x - 2 sin x (1 - cos²x)]
= 18/[(1 + 2 tan x)² cos²x] (cos³x - 2 sin³x)
= 18/[(1 + 2 tan x)² cos²x] (cos x - ∛2 sin x) (cos²x + ∛2 cos x sin x + ∛4 sin²x)

Note that 18/[(1 + 2 tan x)² cos²x] > 0
and cos²x + ∛2 cos x sin x + ∛4 sin²x > 0

Therefore, dy/dx = 0
⇒ cos x - ∛2 sin x = 0
⇒ 1 - ∛2 tan x = 0
⇒ tan x = 1/∛2
⇒ x = nπ + tan⁻¹(1/∛2) where n is an integer.

One can check that
x = 2nπ + tan⁻¹(1/∛2) refers to maximum y and
x = (2n + 1)π + tan⁻¹(1/∛2) refers to minimum y.

local maximum y is 18 ³√2 / (4＋³√2＋2 ³√4)

2015-07-29 09:26:37 補充：
知足 老師講得啱，我漏咗"開方"。