Question 182631: A wire 12 inches long is cut into two pieces. One piece is bent to form a square. The other is bent to form a rectangle which is 1 inch longer than it is wide. How long is each piece if the sum or the area is a minimum?
All I know is that it may help to use the equation 
Answer by kev82(151)   (Show Source):
You can put this solution on YOUR website! So the wire is cut into to pieces, some of it is used to make a rectangle, the rest a square. Lets's say the amount used for the rectangle is x. This means the amount used for the square is 12-x.
Lets's consider the square, it has perimeter 12-x, a square has 4 sides, each the same length. So the length of one side of this square will be 3-x/4. So the area of the square will be (3-x/4)^2.
The rectangle has perimeter x, the formula for perimeter is 2*(w+l), but we know from the question that l=w+1. So 2*(w+w+1) = x, 4w+2 = x, w =(x-2)/4. The area of a rectangle is w*l, so in this case it is. (x-2)(x+2)/16.
So the formula for the total area is
(3-x/4)^2 + (x-2)(x+2)/16
I don't like fractions very much, so I am going to simplify this by letting y=x/4
(3-y)^2 + y^2 - 1/4
Looking at this, you can see the dominating term is a positive y^2 (expand the first brackets if you are not convinced). This means the function looks approximately like y^2, which has a minimum at it's turning point. If we were looking for a maximum, then you wouldn't differentiate as the maximum of a y^2 graph is at its endpoints. Of course if it was dominated by negative y^2 the the max would be at the turning point and the min at the end points. Anyway, we have to differentiate to find this turning point.
Remember I'm differentiating wrt x, and dy/dx=1/4. (Use chain rule)
-2*(3-y)/4 + 2y/4 = 0
4y=6
y = 3/2
but remember y=x/4, so x=6. That means the minimum area is when you give exactly half of the string to each.
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