Optimization

It just means that we want to maximize the area of trapezium.

A
=(10+20cosθ+10)*(10sinθ)/2
=100(1+cosθ)(sinθ)

Take the derivative,
dA/dθ=100(cosθ+sinθ(-sinθ)+(cosθ)(cosθ))
Set dA/dθ=0
cosθ+(cosθ)^2=(sinθ)^2
secθ+1=(tanθ)^2
secθ+1+1=1+(tanθ)^2=(secθ)^2
(secθ)^2-secθ-2=0
(secθ-2)(secθ+1)=0
secθ=2 or secθ=-1
θ=60 or θ=180 (rejected)
So the angle θ should be 60 degree so that the gutter will carry the maximum amount of water.


2009-11-18 21:06:53 補充：
I agree with you 石石

how do you get secθ+1=(tanθ)^2?

theta=pi/3

2009-11-19 22:18:18 補充：
Consider 1/cosθ=secθ, tanθ=sinθ/cosθ,

cosθ+(cosθ)^2=(sinθ)^2
Dividing both sides with (cosθ)^2 will yields:
(1/cosθ)+1=(sinθ/cosθ)^2
secθ+1=(tanθ)^2

let theta be #

The gutter will carry max. aount of water when the area of bended parts are max.

Let area of 1 bended area be A
A=0.5(10sin#)(10cos#)=50sin#cos#=25sin2#.
dA/d#=50cos2#
let dA/d#=0
cos2#=0
2#=90* (0*<#<90*)
#=45*

So, the gutter will carry the maximum amount of water when #=45*