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 Fenced In (Posted on 2004-06-18)
A farmer wishes to enclose the maximum possible area with 100 meters of fence. The pasture is bordered by a straight cliff, which may be used as part of the fence. What is the maximum area that can be enclosed?

 See The Solution Submitted by SilverKnight Rating: 3.7500 (4 votes)

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 Where did my math go wrong? | Comment 4 of 15 |

[*] this post was modified to correct formatting problems in the formulas and correct the Basic program to use radians for the SIN() function...

Area of a segment is given as:

`Area = 1/2•R^2•(PI/180•THETA-Sin(THETA) in degrees`

arc length is given as:

`Arc Length = (THETA•PI)/180•R  in degrees`
`100 = (THETA•PI•R)/18018000 = THETA•PI•Rr = 18000/(THETA•PI)`

If I plug 18000/(THETA•PI) back intot he original formula for the area of a segment in place of R I get:

`Area = 1/2•(18000/(THETA•PI))^2•(PI/180•THETA-Sin(THETA) <- assuming that SIN() uses degrees...`

I then use the following program to check each degree from 1 to 180 to see which gives the maximum area for the segment:

`Dim I, Degrees As IntegerDim Area, TopArea, Pi As Double Pi = 3.1415926Degrees = 0TopArea = 0 For I = 1 To 180    Area = (1 / 2) * (18000 / (I * Pi)) ^ 2 * (Pi / 180 * I - Sin(Pi * I / 180))    If Area > TopArea Then        TopArea = Area        Degrees = I    End IfNext Text1.Text = DegreesText2.Text = TopArea`

This gives me the maximum area of 848006 m² and a value of degrees of 4. [*]

What did I do wrong?  Did I forget to convert between degrees and radians somewhere?[*]

[*]  The corrected program returns a value of 180° for Theta and a total area for the segment as 1591.5494...

Edited on June 18, 2004, 1:14 pm

Edited on June 18, 2004, 1:34 pm
 Posted by Erik O. on 2004-06-18 10:51:31

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