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.
SydneyRoverP6B said: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.
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
SydneyRoverP6B said: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.