Maximum Area of an ArcDate: 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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