optimization problems

[jeca86]

a boat leaves a dock at 2:00pm and travels due south at a speed of 20km/h. another boat has been heading due east at 15km/h and reaches the same dock at 3:00pm. at what time were the two boats closests together?



a rain gutter is to be constructed from a metal sheet of width 30cm by bending up one-third of the sheet on each side through an angle theta. HOw should theta be chosen so that the gutter will carry the maximum amount of water?

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[jeca86]

picture didn't come out right

 

[soroban]

Hello, jeca86!

A rain gutter is to be constructed from a metal sheet of width 30cm
by bending up one-third of the sheet on each side through an angle \(\displaystyle \theta.\)
How should \(\displaystyle \theta\) be chosen so that the gutter will carry the maximum amount of water?

     10cosθ    10    10cosθ
     * . . . . . . . . . *
     :\   :         :   /:             The cross-section of the gutter
     : \10:         :10/ :                    is a trapezoid.
     :  \ :         : /  : 10sinθ           The lower base is 10.
     :  θ\:         :/θ  :               The upper base is 10 + 20cosθ.
     :....* - - - - *.....                   The height is 10sinθ.
               10   10cosθ

The area of a trapezoid is: . \(\displaystyle A\;=\;\frac{h}{2}(b_1\,+\,b_2)\)

So we have: .\(\displaystyle A\;=\;\frac{10\sin\theta}{2}[10\;+\,(10\,+\,20\cos\theta)]\)

. . which simplifies to: .\(\displaystyle A\;=\;100\cdot\sin\theta(1\,+\,\cos\theta)\)

And that is the function we must maximize . . .

 

[mammothrob]

heres a tip.

draw a right triangle and find length of the legs in terms of time.

you can find how far the boat thats leaving is... theres one leg
[distance in miles * (t)]
and because you know how the other boat is closing in one dock you can get something like [hours - distnace in miles(t)] or something similar, make that your other leg. the distance between the two is your hypotenuse of the right triangle.

use the distnace formula to make a function

d = Sqrt. ([hours - distance in miles * (t)]^2 + [distance in miles * (t)]^2)

when you simplify use what is under the radical... dont get rid of the radical just take what you simplified under it and make it a function... use the derivative and find a critical numbers. There will probably only be one though and it should be a min because the distnace will only increase as your fuction of distance gets largers.

Its late, so I hope that was understandable. Your set up may look a tiny different than mine though... just think about the problem.