You are in possession of a 4 metre length of wire for making a circle and a square. How should the wire be distributed between the two shapes in order maximise the sum of the enclosed areas?
Ron.
Ron.
A Circle and a Square
- Thread starter SydneyRoverP6B
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No takers... 
Richard and Darth,....what are your thoughts gents?
Ron.
Richard and Darth,....what are your thoughts gents?
Ron.
darth sidious
New Member
SydneyRoverP6B said:
Unless I am very much mistaken here, the maximum possible enclosed area would be to forget the square, and just make a 4m circumference circle. That gives an area of 4/pi m^2 (or 1.273239545 m^2)
4 = 2*pi*r so r =2/pi, r^2 = 4/(pi ^2), so Area = pi * r^2 = 4/pi
You can differentiate an expression of total area with respect to the length of each side of the square (which I denote by "x"), but that only gives a minimum value this time... The minimum is when the :-
length of each side of the square is 1/[1 + pi/4] (= 0.5600992)m (560.0992mm), giving an area 0.3137111 m^2 (313711.1 mm^2) the circumference of the circle is 4*pi/[4+pi] (=1.7596034)m, the diameter [4/(4+pi)] (=0.5600992) m, radius [2/(4+pi)] (=0.2800496)m (280.0496mm), and area is then 4*pi/[{4+pi}^2] (=0.2463881) m^2
The total area is then 0.5600992 m^2 (560099.2 mm^2)
That is the minimum value I can get, and Excel spreadsheet seems to concur.
Of course, if I am mistaken, please do feel free to show me the error of my ways!
Yes, I would agree
The area of a circle with a 4M diameter is 1.273M^2
A square with 1 metre sides is 1M^2
The smallest area would be 0.56M^2 with a square with sides of 0.56M (2.24 perimeter) and a circle with a diamter of .56M (circumference of 1.76M).
Or very close to it
The area of a circle with a 4M diameter is 1.273M^2
A square with 1 metre sides is 1M^2
The smallest area would be 0.56M^2 with a square with sides of 0.56M (2.24 perimeter) and a circle with a diamter of .56M (circumference of 1.76M).
Or very close to it
darth sidious
New Member
quattro said:Yes, I would agree
The area of a circle with a 4M diameter is 1.273M^2
A square with 1 metre sides is 1M^2
The smallest area would be 0.56M^2 with a square with sides of 0.56M (2.24 perimeter) and a circle with a diamter of .56M (circumference of 1.76M).
Or very close to it
Wow!!! I am honoured that Richard agrees with me!
Hello Darth and Richard,
Well done gents!.... Yes indeed that is the correct answer.
Deriving an expression for the total area in terms of x and then differentiating, equating to zero and solving for the critical point gives a solution of x = 4PI/(PI + 4). Testing indeed does reveal a minimum value within the closed interval....that is to say a local minimum value. So testing the boundaries of the closed interval within the area expression will give a global perspective and this shows that when x = 4 the area is a maximum, and this means that the square is not included.
Ron.
Well done gents!....
Yes indeed that is the correct answer. Deriving an expression for the total area in terms of x and then differentiating, equating to zero and solving for the critical point gives a solution of x = 4PI/(PI + 4). Testing indeed does reveal a minimum value within the closed interval....that is to say a local minimum value. So testing the boundaries of the closed interval within the area expression will give a global perspective and this shows that when x = 4 the area is a maximum, and this means that the square is not included.
Ron.
darth sidious
New Member
SydneyRoverP6B said:Hello Darth and Richard,
Well done gents!....
Deriving an expression for the total area in terms of x and then differentiating, equating to zero and solving for the critical point gives a solution of x = 4PI/(PI + 4). Testing indeed does reveal a minimum value within the closed interval....that is to say a local minimum value. So testing the boundaries of the closed interval within the area expression will give a global perspective and this shows that when x = 4 the area is a maximum, and this means that the square is not included.
Ron.
Well done both of us!!