Question 194617: a 32inch 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 areas of the square and the rectangle?
Answer by solver91311(24713)   (Show Source):
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Let x represent the width of the rectangle. Then x + 2 represents the length of the rectangle.
That means that the perimeter of the rectangle is  , and the perimeter of the square is then  = 28 - 4x) , so the side of the square must measure 
The area of the rectangle is then  and the area of the square is ^2 = x^2 - 14x + 49) and the sum of these two areas gives the total area function with respect to the width of the rectangle:
 = 2x^2 - 12x + 49)
for which you want to find a minimum value.
Algebra Solution:
This function represents a parabola opening upwards because of the positive lead coefficient. Since the parabola opens upward the value of the function at the vertex will represent a minimum values. The x-coordinate of the vertex for any parabola of the form  = ax^2 + bx + c) is given by  . For this situation:
}{2(2)} = 3)
But before we do anything with that, we need to make sure that 3 makes sense as a value for the width. The domain of the total area function, in that it is a polynomial function with integer coefficients, is all real numbers. But for this situation, the actual domain is somewhat restricted. Certainly the width of the rectangle cannot be less than zero, so that gives us a practical bottom end to the area domain. On the other end, if the width were greater than 7, then the length would be greater than 9 and the perimeter would then be larger than the 32 inches we started with. Hence, the practical domain of the total area function is:
In that 3 is contained in that interval, 3 is an acceptable solution. The value of the function at 3 is:
 = 2(3)^2 - 12(3) + 49 = 31)
Which makes sense because if the width of the rectangle is 3, then the length is 5 making the area of the rectangle 15. It also makes the perimeter of the rectangle 16, so the perimeter of the square must also be 16, the side 4, and the area 16. So the total area is 31. This also gives us the answer to the question posed, namely you cut the wire exactly in half -- 2 pieces each 16 inches long.
Calculus Solution:
The function is continuous and twice differentiable over its entire domain, so a local extrema can be found wherever the first derivative is equal to zero and that extreme point will be a minimum if the second derivative is positive at that point.
= 4x - 12)
Hence the function has an extreme point at ))
 = 4) , hence positive for all x.
Hence is a local minimum.
Therefore width = 3, length = 5, perimeter of rectangle = 16, i.e., cut the wire in half.
John
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