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1.已知一個長方形的長為4,闊為3,若長增加x單位,同時闊減少x/2單位,問當x取何值時該長方形的面積為最大?又長方形的面積最大為多少?

2007-04-25 20:17:47 · 2 個解答 · 發問者 yvonne 1 in 科學 數學

請問如果我咁做錯在那裏呢?thx
A=(4+x)(3-x/2)
=12+x-(x^2)/2
=24+2x-x^2
= -x^2+2x+24
= -(x^2+2x+1)+26
= -(x+1)^2+26
when x= -1 A(max)=26

2007-04-26 12:07:36 · update #1

2 個解答

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

2007-04-25 20:38:29 · answer #1 · answered by §◎◎§ ┘咪go佐敦㊣ 5 · 0 0

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

2007-04-25 20:51:32 · answer #2 · answered by 老爺子 7 · 0 0

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