A piece of wire 60cm long is to be cut into two pieces. One piece will be bent to form a circle and other piece will be bent to form a square.

Question

Didn’t finish the entire question but here it is: Find the length of the two pieces that will cause the sum of the areas of the circle and square to be a minimum.

Answer 1

This is typically presented as a problem in calculus, but can be solved using algebra because the Total Area function results in a upward opening parabola. Locate the vertex and the x-coordinate represents one of the pieces of the string.

Start by letting x = length of one piece, and 60-x the length of the other piece. One of those pieces will be folded into a square the other a circle. For this discussion the piece represented by x will turn into a square.

SQUARE

If perimeter is x, the each side is x/4.

Therefore area of square is x²/16,

CIRCLE

If circumference is 60-x then the radius is (60-x)/2π.

Therefore the area of the square is π(60-x)²/(2π)²=(60-x)²/4π.

Total Area Function

T(x) = x²/16 + (60-x)²/4π

Get that function in the form T(x) = ax²+bx+c.

The axis of symmetry will be at x = -b/2a and that value will be the x-coordinate of the vertex, which in turn will be the length of the 60 cm string that should be the perimeter of the square.

Both a and b will have π in their respective representations.

After you have tried all of the above you can check your work here:

desmos.com/calculator/0wo1w6qhwb