I think you are supposed to use quadratics, but I don't know how.
A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the other into a circle. How can the wire be cut so that the area enclosed by the two shapes is minimized? Give your answer to the nearest tenth.
+128399 Let x be the part of the wire reserved for the circle
This must be the perimeter of the circle
Perimeter = 2pi*r
x =2pi* r
[ x / (2pi) ] = r
So the area of the circle = pi* r^2 = pi [ x / (2pi)]^2 = [x^2] / [ 4pi ]
The part reserved for the squre is just the perimeter of the square = [ 10- x ]
The side of the square = perimeter /4 = [ 10 - x ] /4
Area of the square = side^2 = [ 10 - x ]^2 / 16
So....the total area of the circle and square , A, can be expressed as
A = x^2 / [ 4pi ] + [ 10 - x ]^2 / 16 simplify
A = x^2 / [ 4pi ] + [ x^2 - 20x + 100] /16
A = x^2 / [4pi] + (1/16)x^2 - (5/4)x + 25/4
This is most easily done with Calculus.....I'll graph the curve to show you that the value for x obtained is the correct one
Take the derivative of A and set it to 0
2x /(4pi) + (2/16)x - 5/4 = 0
x / [ 2pi] + x /8 = 5/4
x [ 1/[ 2pi] + 1/8] = 5/4
x [ 1/(2pi) + 1/8 ] = 1.25
x = 1.25 / [ 1/(2pi) + 1/8] ≈ 4.399 ≈ 4.4
Cut the wire at about 4.4 inches.....this is the part reserved for the circle
So the part reserved for the square =10 - 4.4 = 5.6 in
Here is a graph showing that the value for x is correct :
https://www.desmos.com/calculator/d8t5qos9c3
