a wire is 36 m long is cut into two pieces , each piece is bent to form a rectangle which is 1 centimeter longer than its width. how long should each piece be to minimize the sum of the areas of the two rectangles?
seems like I saw this one go by the other day, but here it is again:
If the two pieces are x and y in length, then the rectangles have dimensions
(x-2)/4 by (x-2)/4 + 1 or
(x-2)/4 by (x+2)/4
and
(y-2)/4 by (y+2)/4
But, y = 36-x, so the rectangles are
(x-2)/4 by (x+2)/4
(34-x)/4 by (38-x)/4
and the sum of their areas is
(x-2)(x+2)/16 + (34-x)(38-x)/16
= (x^2-36x+644)/8
= 1/8 [(x-18)^2 + 320]
This is a parabola with vertex at (18,40), where the area is a minimum.
avoiding some of those fractions
let the sides of one rectangle be x and x+1
and the other, y and y+1
4x+2 + 4y+2 = 36
x+y=8
y = 8-x
area = A = x(x+1) + y(y+1)
= x^2 + x + y^2 + y
= x^2 + x + 64-16x+x^2 + 8-x
= 2x^2 - 16x + 72
Quadratic opening up, so a minimum
dA/dx = 4x - 16
= 0 for a min
x = 4
then y = 4
The two rectangles are the same size, both 4 by 5
so we need 2(4+5) or 18 m for each
for a min area of 4(5) + 4(5) or 40 m^2
for a min
Well, well, well, let's find the right length for those rectangles! To minimize the sum of their areas, we need to dive into some mathematical clownery.
Let's call the width of each rectangle x. Now, since each rectangle is 1 centimeter longer than its width, the length of each rectangle would be (x + 1). Keep in mind that we have two pieces formed from the wire.
The perimeter of each rectangle is equal to twice the sum of the length and width. So, for each rectangle, we have 2(x + (x + 1)) = 4x + 2.
Since we are given that the wire has a total length of 36 meters, we can write the equation: 2(4x + 2) = 36. Solving this equation will give us the value of x, and then we can calculate the lengths of the two pieces.
But since we're clowns here, why not have some fun and solve it step by step? *Drumroll*
2(4x + 2) = 36
8x + 4 = 36
8x = 36 - 4
8x = 32
x = 32/8
x = 4
Haha! Our clown noses have led us to discover that x equals 4! Now, since x represents the width of each rectangle, both pieces should have a width of 4 centimeters.
And since each length is 1 centimeter longer than its width, each piece should have a length of (4 + 1) = 5 centimeters.
So, to minimize the sum of the areas of the two rectangles, each piece should be 4 centimeters wide and 5 centimeters long.
Hope that brings a smile to your mathematical journey!
To minimize the sum of the areas of the two rectangles, we need to express the sum of the areas in terms of a single variable and then find the minimum of that expression using calculus.
Let's start by defining our variables. Let the length of the first piece be x meters, and let the length of the second piece be (36 - x) meters, since the wire is 36 meters long and we're dividing it into two pieces.
Given that each piece is bent to form a rectangle with width w and length (w + 1) centimeters, we can write the following equations for the perimeters of the two rectangles:
For the first piece:
2w + 2(w + 1) = x
For the second piece:
2w + 2(w + 1) = (36 - x)
Simplifying these equations, we get:
4w + 2 = x
4w + 2 = 36 - x
Now, let's express the areas of the two rectangles in terms of the width w:
For the first rectangle:
Area1 = w * (w + 1)
For the second rectangle:
Area2 = w * (w + 1)
The sum of the areas of the two rectangles is:
Sum of Areas = w * (w + 1) + w * (w + 1)
Simplifying this expression, we get:
Sum of Areas = 2w^2 + 2w
To find the minimum of this expression, we can take the derivative with respect to w and set it equal to zero:
d(Sum of Areas)/dw = 4w + 2 = 0
Solving this equation, we find:
w = -0.5
However, a negative value for the width doesn't make sense in this context. So, let's consider the constraints of the problem. The rectangles must have positive widths and lengths, so we have the following additional constraints:
w > 0
w + 1 > 0
36 - x > 0
Using these constraints, we can determine the valid range for w. Since the smallest value for w is 0.5 (which corresponds to a width of 0.5 cm), and w cannot go beyond (36 - x) due to the constraint 36 - x > 0, we have:
0.5 ≤ w ≤ 36 - x
Now, we can evaluate the expression for the sum of the areas at the endpoints of this range:
When w = 0.5:
Sum of Areas = 2(0.5)^2 + 2(0.5) = 1.5
When w = 36 - x:
Sum of Areas = 2(36 - x)^2 + 2(36 - x) = 144 - 4x + 2x^2
To determine the minimum value of the sum of areas, we can compare these two values and choose the smaller one.
So, the final step is to compare:
1.5 and 144 - 4x + 2x^2
To find the minimum value of the second expression, we can take its derivative with respect to x, set it equal to zero, and solve for x.
d(144 - 4x + 2x^2)/dx = -4 + 4x = 0
Solving this equation, we find:
x = 1
We can verify that this point corresponds to a minimum by taking the second derivative and ensuring it is positive:
d^2(144 - 4x + 2x^2)/dx^2 = 4 > 0
Therefore, x = 1 is the value that minimizes the sum of the areas of the two rectangles.
To find the lengths of the two pieces, we substitute x = 1 into our earlier equations:
For the first piece:
4w + 2 = 1
4w = -1 (which is not valid)
For the second piece:
4w + 2 = 36 - 1
4w = 35
w = 8.75
So, the length of the first piece is approximately 1 meter, and the length of the second piece is approximately 35 meters.