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### Maximum Area of an Arc

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Date: 12/4/95 at 22:40:49
From: Anonymous
Subject: maximization in calculus (using trig)

Problem:  A string of length 25cm is to be formed into the arc of a
circle so that the area of the segment formed will be a maximum.

Analysis:  I get the area function to be A= (625/8x^2)(2x-sin2x), where
2x is the central angle.  Unfortunately, that function has no maximum or
minimum.  Further, it is "obvious" that 2x must equal Pi radians so that
the segment is really a semicircle.  The correct answer must be that the
radius for maximum area is 25/Pi.

I can't get an algebraic solution!  Can you?
```

```
Date: 5/30/96 at 14:55:58
From: Doctor Anthony
Subject: Re: maximization in calculus (using trig)

Let x = angle at the centre of the circle subtended by the arc of 25
cms.

So rx = 25 and x = 25/r
The area of the segment =(1/2)r^2.x - (1/2)r^2.sin(x)
Substitute for x (=25/r)

Area = (1/2)r^2.(25/r) - (1/2)r^2.sin(25/r)
= (25/2)r - (1/2)r^2.sin(25/r)

Differentiate with respect to r

dA/dr = 25/2 - (1/2).2r.sin(25/r) - (1/2)r^2.cos(25/r).(-25/r^2)

= 25/2 - r.sin(25/r) + (25/2).cos(25/r)

= (25/2)(1 + cos(25/r)) - r.sin(25/r)

Go to half angles to factorize.

= (25/2).2cos^2(25/2r) - 2r.sin(25/2r).cos(25/2r)

= 2cos(25/2r){(25/2)cos(25/2r) - rsin(25/2r)}

So either cos(25/2r) = 0 i.e.  25/2r = pi/2

r = 25/pi

or r.sin(25/2r) = (25/2)cos(25/2r)
tan(25/2r) = (25/2r) requiring 25/2r = 0.  This is not a possible
solution.

So maximum area occurs when  r = 25/pi

-Doctor Anthony,  The Math Forum

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Associated Topics:
High School Calculus

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