sadnessfung

Q14  如下圖  感激

等運到

Are you sure this is a M1 question? I think it is a typical question that may appear in the compulsory part.

Let AE = k.
You can express the area of CEF as a quadratic function of k.
Then you can calculate the maximum area by using the formula or by completing the square.

trt24418132

Let AE=k cm ,AF=2k cm, area of triangle CEF=y cm2
triangle CEF=y=a^(2)-2k^(2)(1/2)-(a-2k)(a)(1/2)-(a-k)(a)(1/2)
                          y =a^(2)-k^(2)-a^(2)+(3ak)/2
                          dy/dk=-2k+(3a)/2=0
                          k=(3a)/4
                          y=(9a^(2))/16+(9a)/8
so, the maximum area is 0.5625a^(2)+1.125a cm2

sadnessfung

引用:

原帖由 trt24418132 於 2015-8-26 05:01 PM 發表
Let AE=k cm ,AF=2k cm, area of triangle CEF=y cm2
triangle CEF=y=a^(2)-2k^(2)(1/2)-(a-2k)(a)(1/2)-(a-k)(a)(1/2)
                          y =a^(2)-k^(2)-a^(2)+(3ak)/2
                          dy/d ...

the answer is 9/16a^2 cm^2.....


Thank you so much!  I had differentiated the"a" instead of k...  I didn't realize a is constant   and k is the variable...

That's the reason why I failed to get the answer...  HAHA!  thank you !

[ 本帖最後由 sadnessfung 於 2015-8-26 05:38 PM 編輯 ]

sadnessfung

Maybe it can be solved by completing square method...  yet it is a question in the m1 book... so I would like to use the differentiation to solve it....

I got some careless mistakes though I knew AE:AF =1k:2k   ...  

anyway! thank you

膠庄

It seems that you didn't notice the constraint of 0<2k<a, i.e. 0<k<a/2.
This is because "E and F are two points of AD and AB respectively".
Therefore 0<AF<AB, i.e. 0<2k<a.
As a result, k cannot be 3a/4.

The correct answer should possibly be nearly (a^2)/2 with k extremely near a/2 and 2k extremely near a.

Hope this helps

peterkcc2015

As you say, let
﹒ ﹒ ﹒ AE = k; 0 ≦ k ≦ a
﹒ ﹒ ﹒ AF = 2k; 0 ≦ 2k ≦ a
﹒ ﹒ ﹒ Area of ΔCEF = A(k)
first of all
﹒ ﹒ ﹒ 0 ≦ k ≦ a and 0 ≦ 2k ≦ a
⇒ ﹒ ﹒ 0 ≦ k ≦ a/2
secondly
﹒ ﹒ ﹒ A(k) = a² - (k)(2k)/2 - (a - 2k)(a)/2 - (a - k)(a)/2
﹒ ﹒ ﹒ = 3ak/2 - k²
set the 1st derivative to zero
﹒ ﹒ ﹒ d[A(k)]/dk = 3a/2 - 2k = 0
﹒ ﹒ ﹒ k = 3a/4 (rejected as k ≦ a/2)
fortunately
﹒ ﹒ ﹒ d[A(k)]/dk > 0 for 0 ≦ k ≦ a/2
which means A(k) is increasing and
﹒ ﹒ ﹒ max A(k) = A(a/2) = (3a/2)(a/2) - (a/2)²
﹒ ﹒ ﹒ = a²/2 cm² §


9a²/16 cm²的答案有誤，請再查証。

[ 本帖最後由 peterkcc2015 於 2015-8-29 04:47 PM 編輯 ]

Patchouli

從F向右畫一水平線至EC, 交點為G,

三角形CEF面積=(FG*AF+FG*FB)/2=FG*AB/2=a*FG/2
當F點向下移, E點向右移, G點同樣, 三角形CEF面積增加
當F點到達A點, G點到達C點, FG為最大值, FG=AC=a
所以三角形CEF的最大面積=a^2/2 cm^2

不微分也能看得出