SOLUTION: The question in the book reads "A piece of wire 20 inches long is to be cut into two pieces, one of which will be bent into a circle and the other into a square. How long should ea

Question 106776This question is from textbook College Algebra
: The question in the book reads "A piece of wire 20 inches long is to be cut into two pieces, one of which will be bent into a circle and the other into a square. How long should each piece be to minimize the sum of the areas?". So one section of the wire could be represented as "x" and the other piece as "20-x". That's all I can come up with. Any help would be greatly appreciated!! This question is from textbook College Algebra

Found 2 solutions by scott8148, TP:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
good start ... let x=circle, so 20-x=square

x is circumference, so radius is x/2pi and area is x^2/4pi

20-x is perimeter, so side is (20-x)/4 and area is ((20-x)^2)/16 ... (400-40x+x^20/16

sum of areas is ((4+pi)/16)x^2-2.5x+25

minimum is on the axis of symmetry (x=-b/2a)
Answer by TP(29) (Show Source): You can put this solution on YOUR website!
The total length is 20 inches.
Let the two lengths be x and 20-x inches
Using calculus:
The area of the square is (x/4)^2 and the area of the circle is pi*radius^2.
Now the circumference = 2*pi*r and C=20-x.
Hence r=(20-x)/(2*pi)= 10/pi-x/2pi
So total area is A=(x/4)^2+pi*(10/pi-x/2pi)^2=(x^2)/16+pi*(100/(pi)^2+x^2/4pi^2-10x/pi^2)
A= x^2/16 + 100/pi + x^2/4pi - 10x/pi
Maximum or minimum values are given when dA/dx=0 and so:
2x/16+2x/4pi - 10/pi = 0
x/8+x/2pi - 10/pi = 0
Multiply by 8pi
x*pi + 4x - 80 = 0 = 0
x*(pi + 4) - 80 = 0
x = 80/(pi + 4)
x = 80/(22/7 + 28/7) = 80/(50/7) = (8*7/5 = 56/5 = 11.2
So the lengths are 11.2 inches and 8.8 inches ANS

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