[中六] 文達附加數 3題

我諗第２題４０分成兩部分，一部分唔駛平方，另一數先要平方吧～
兩個數仲要乘埋添，即係咁:
設兩數為ｘ和ｙ。
ｘ＋ｙ＝４０
要使 (ｘ)(ｙ^２) 最大.... ??

中6都要讀附加數？!

我不知可否用英文吧=.=..

1.
let the two numbers is x and 100-x..

積y
=x(100-x)
=-x^2 + 100x

dy/dx= -2x+100 = 0
x=50..
so the maximum is 2500 when x = 50

2.
let the two number is x and 40-x
平方之乘積 y
= x^2 * (40-x)^2
=(40x-x^2) ^2

dy/dx=2(40x-x^2)(40-2x) =0
x= 0(rej.) or 40 (rej.)or 20

so the answer is 20
it should show when x=20..it is the maximum..
you can show it by second derivative..

3.
let the length is x and 36-x..

Area y
= x(36-x)
= 36x-x^2

dy/dx=36-2x =0
x=18..

because there is no minimum..
so when x=18...the rectangle has the largest area..
also..you can show it by second derivative...


actually...the above question can be solved by same method..
if you do not know what the calculus is..
you can use completing square to find the vertex for quadratic eqt..

1
令兩數為x,y
x+y=100﹐故y=100-x
考慮xy=x(100-x)=-x^2+100x
令d/dx(-x^2+100x)=0
-2x+100=0
x=50
所以x=y=50
2
令兩數為x,y
x+y=40﹐故y=40-x
考慮x^2+y^2=x^2+(40-x)^2=2x^2-80x+1600
令d/dx(2x^2-80x+1600)=0
4x-80=0
x=20
所以x=y=20
3
令兩數為x,y
2(x+y)=72﹐故y=36-x
考慮xy=x(36-x)=-x^2+36x
令d/dx(-x^2+36x)=0
2x=36
x=18
所以x=y=18
其尺寸為長18 cm 闊18 cm

2008-12-28 19:39:09 補充：
我漏做了用2階導數檢查﹐不過因為這些是物理/應用題﹐所以其實一定得到最大值