SOLUTION: a 32 inch piece of wire is to be divided into 2 pieces. one piece is going to be bent to form a square. the other piece is going to be bent into a rectangle which is 2 inches longe

Question 194402: a 32 inch piece of wire is to be divided into 2 pieces. one piece is going to be bent to form a square. the other piece is going to be bent into a rectangle which is 2 inches longer than it is wide. how long should each piece of wire be to minimize the sum of the ares of the square and the rectangle?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
a 32 inch piece of wire is to be divided into 2 pieces.
one piece is going to be bent to form a square.
the other piece is going to be bent into a rectangle which is 2 inches longer than it is wide.
how long should each piece of wire be to minimize the sum of the ares of the square and the rectangle?
:
let s = side of the square
:
Let x = width of the rectangle
and
(x+2) = length of the rectangle
:
Using wire length equation, find the side of the square in terms of x
4s + 2x + 2(x+2) = 32
4s + 2x + 2x + 4 = 32
4s + 4x = 32 - 4
4s + 4x = 28
simplify, divide by 4
s + x = 7
s = (7-x)
:
Area equation
A = square area + rectangle area
A = s^2 + x(x+2)
Substitute (7-x) for s
A = (7-x)^2 + x(x+2)
A = 49 - 14x + x^2 + x^2 + 2x
Arrange as a quadratic equation
A = 2x^2 - 12x + 49
:
The axis of symmetry will give the value of x for minimum area (x = )
a = 2; b = -12
x =
x =
x = 3 inches
:
Length of rectangle wire:
2x + 2(x+2) =
2(3) + 2(3+2) = 16 inches
:
Length of square wire:
4(7-x) =
4(7-3) = 16 inches
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