SOLUTION: A piece of wire 6 metres long is cut into 2 parts. One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth. Find the lengths of the

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A piece of wire 6 metres long is cut into 2 parts. One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth. Find the lengths of the       Log On


   

Question 1012461: A piece of wire 6 metres long is cut into 2 parts. One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth. Find the lengths of the 2 parts if the sum of the two areas is a minimum.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A piece of wire 6 metres long is cut into 2 parts.
One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth.
Find the lengths of the 2 parts if the sum of the two areas is a minimum.
:
let s = the side of the square
let L = the length of the rectangle
let W = the width
therefore the two perimeters = 6
4s + 2L + 2W = 6
Simplify, divide equation by 2
2s + L + W = 3
Given that
L = 3W
therefore
2s + 3W + W = 3
2s + 4W = 3:
Simplify, divide by 2
s + 2W = 1.5
s = -2W+1.5
:
the total area
A = s^2 + LW
Replace s with (-2W+1.5), replace L with 3W
A = (-2W+1.5)^2 + 3W(W)
FOIL
A = 4W^2 - 3W - 3W + 2.25 + 3W^2
Combine like terms
A = 7W^2 - 6W + 2.25
Find the minimum by finding the axis of symmetry, x=-b/(2a)
w = %28-%28-6%29%29%2F%282%2A7%29
W = 6%2F14
W = .42857 is the width for minimum area
Find L
3(.42857) = 1.2857 is the length
Find the length of the portion made into a rectangle (perimeter)
P = 2(1.2857 + 2(.42857)
P = 3.42856m is the portion made into a rectangle
then
6 - 3.42856 = 2.57m is the portion made into a square
:
: