SOLUTION: A 30 cm piece of wire is cut into two pieces. One piece is bent into the shape of a square the other piece is bent into the shape of a rectangle with length to width ratio of 2:1 W

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Question 750169: A 30 cm piece of wire is cut into two pieces. One piece is bent into the shape of a square the other piece is bent into the shape of a rectangle with length to width ratio of 2:1 What are the length's of the two pieces if the sum of the areas of the square and rectangle is a minimum?
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Answer by ankor@dixie-net.com(22740) (Show Source):
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A 30 cm piece of wire is cut into two pieces.
One piece is bent into the shape of a square. the other piece is bent into
the shape of a rectangle with length to width ratio of 2:1
What are the length's of the two pieces if the sum of the areas of the square
and rectangle is a minimum?
:
let x = length of the side of square
let y = length of the width of the rectangle
then
2y = Length of rectangle
So we have
4x + 2(2y) + 2(y) = 30
4x + 6y = 30
6y = 30 - 4x
simplify, divide by 2
3y = 15 - 2x
Divide by 3
y = (5-x)
the total area equation
A = x^2 + (2y*y)
A = x^2 + 2y^2
replace y with (5-x) which is (5-.67x)
A = x^2 + 2(5-.67x)^2
FOIL (5-.67x)(5-.67x)
A = x^2 + 2(25-6.7x+.44x^2)
A = x^2 + 50 - 13.14x + .88x^2
A = 1.88x^2 - 13.14x + 50
Find the minimum by finding the axis of symmetry using x = -b/(2a)
x =
x =
x = 3.5 cm is the side of the square for minimum area, therefore
4(3.5) = 14 cm is piece made into a square
Find the piece made into the rectangle
y = 5 - .67(3.5)
y = 2.66 cm is the width of the rectangle, therefore
6(2.66) ~ 16 cm is the piece made into the rectangle
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