SOLUTION: Minimizing area. A 46-in piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the

Algebra ->  Test -> SOLUTION: Minimizing area. A 46-in piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the      Log On


   

Question 746002: Minimizing area. A 46-in piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the sum of the areas is minimum?
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Let L = the length of the string = 46 in
Let x = the length of one piece
Then L-x = the length of the remaining piece
If the circle is formed from the 1st piece, the radius of the circle will be x/(2*pi)
The area of the circle will be Ac = pi*(x/(2*pi))^2 = x^2/(4*pi)
The perimeter of the square = L-x, so each side s = (L-x)/4
The area of the square will be s^2 = (L-x)^2/16
To minimize the sum of the areas, we take the derivative of the sum and set = 0
If you perform these operations and solve for x in terms of L, you should get x = pi/(4+pi)*L
So the areas are minimized if the cut is made at about 0.44*46 = 20.24 in.