面積問題..............急................謝謝!!

Area A=(4+x)(3-x/2)=12+3x-2x-(x^2)/2
A=12+x-(x^2)/2
dA/dx=1-x

A is maximum when dA/dx=0
therefore when x=1, A=maximum or minimum

as d(dA/dx)/dx=-1<0

therefore dA/dx=0 at x=1 is maximum

the maximum Area is A=12+(1)+(1)^2/2=13.5

2007-04-26 01:06:09 補充：
the maximum Area is A=12 (1)-(1)^2/2=12.5

2007-04-26 01:12:08 補充：
[小孩子]的答案較我的答案易明,好野;而我的答案是A-level pure maths 的標準答法,有時數學真是殊途同歸

2007-04-26 01:13:50 補充：
the maximum Area is A=12 (1)-(1)^2/2=12.5

2007-04-26 01:14:35 補充：
the maximum Area is A=12 (1) - (1)^2/2=12.5

2007-04-26 01:15:37 補充：
the maximum Area is 12 1-(1)^2/2=12.5

2007-04-26 01:19:35 補充：
唔知點解個 "add" 符號出唔到,總之答案是12 "add" 1-(1)^2/2=12.5

面積
= (4 + x) (3 - x/2)
= (4 + x) (6 - x) / 2
= [24 + 2x - x^2] / 2
= [25 -1 + 2x -x^2] / 2
= [25 - (1 - 2x + x^2)] / 2
= [25 - (1 - x)^2] / 2

(1 - x)^2 愈小，面積愈大。
(1 - x)^2 最小為 0。
所以面積最大時，(1 - x)^2 = 0
即 x = 1

最大面積
= 25 /2
= 12.5