SOLUTION: a 2 ft. wire is cut into two pieces. one of the pieces, of length x, is bent into a circle. the other piece is bent into a rectangle whose length is twice the size of its width. wh

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Question 211874This question is from textbook just-in-time algebra and trigonometry
: a 2 ft. wire is cut into two pieces. one of the pieces, of length x, is bent into a circle. the other piece is bent into a rectangle whose length is twice the size of its width. what is the total area of the two shapes as a function of x. This question is from textbook just-in-time algebra and trigonometry

Answer by nerdybill(7384) About Me  (Show Source):
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a 2 ft. wire is cut into two pieces. one of the pieces, of length x, is bent into a circle. the other piece is bent into a rectangle whose length is twice the size of its width. what is the total area of the two shapes as a function of x.
.
Let x = circumference of circle
then
2-x = perimeter of rectangle
.
Since for any circle:
circumference = (pi)d = 2(pi)r
then
x = 2(pi)r
r = x/(2(pi))
.
Area of circle:
(pi)r^2
=(pi)[x/(2(pi))]^2]
=(pi)[x^2/(4(pi)^2)]
=x^2/(4(pi))
.
rectangle:
Let w = width
then
2w = length
perimeter = 2(w + 2w)
perimeter = 2(3w)
2-x = 2(3w)
2-x = 6w
(2-x)/6 = w (width)
.
Length:
2w = 2[(2-x)/6] = (2-x)/3 (length)
.
Area of rectangle:
w*L
= (2-x)/6 * (2-x)/3
= (2-x)^2/18
.
Total area:
"area of circle" + "area of rectangle"
= x^2/(4(pi)) + (2-x)^2/18
= 18x^2/(72(pi)) + (4(pi))(2-x)^2/(72(pi))
= [18x^2 + (4(pi))(2-x)^2]/(72(pi))
= [18x^2 + (4(pi))(2-x)(2-x)]/(72(pi))
= [18x^2 + (4(pi))(4-4x+x^2)]/(72(pi))
= [18x^2 + 16(pi) - 16(pi)x + 4(pi)x^2]/(72(pi))
= [18x^2 + 4(pi)x^2 + 16(pi) - 16(pi)x]/(72(pi))
= [2x^2(9 + 2(pi)) + 16(pi)(1-x)]/(72(pi))
= 2[x^2(9 + 2(pi)) + 8(pi)(1-x)]/(72(pi))
= [x^2(9 + 2(pi)) + 8(pi)(1-x)]/(36(pi))
= [x^2(9 + 2(pi)) + 8(pi)(1-x)]/(36(pi))