A wire is 50 inches long. Where should the wire be cut so that it forms 2 pieces one made into a circle and one made into a square such that the area of the circle plus the area of the square in exactly 80 square inches?

Guest Mar 10, 2020

#2**+3 **

"*A wire is 50 inches long. Where should the wire be cut so that it forms 2 pieces one made into a circle and one made into a square such that the area of the circle plus the area of the square in exactly 80 square inches?*"

Hmm!

Alan Mar 10, 2020

#3**0 **

I think your total area should be 800 square inches NOT 80 square inches.

Let x = side length of the square.

50 - x =Circumference of the circle.

[50 - x] / 2pi =Radius of the circle

x^2 + [(50 -x)/(2pi)]^2*pi = 800, solve for x

x =27.567 inches - one side of the square.

50 - 27.567 =22.433 inches - Circumference of the circle.

So that:22.433/2pi =3.57 inches - radius of the circle.

27.567^2 + [3.57^2*pi] =~800 sq.inches.

So, the 50-inch wire should be cut into 2 pieces:

27.567 inches - being one side of the square.

22.433 inches - being the circumference of the circle.

Guest Mar 10, 2020

#4**+3 **

1. If x is the side length of the square then the circumference of the circle is 50 - 4x rather than 50 - x.

2. However, note that the maximum possible area that can be achieved by a piece of wire 50 inches long is when the whole wire is used to make a circle. This results in an area of \(\pi (\frac{50}{2\pi})^2 = 198.94... in^2\) i.e. much less than 800 in^{2}.

Alan
Mar 10, 2020