a wire 36 meter long is cut into two pieces. each piece is bent to form a rectangle which is 1 cm longer than its width. How long should each piece be to minimize the sum of the areas of the two rectangle?
width of each rectangle = x
length of each rectangle = x+1
sum of area = 2x(x+1) = 2x^2 + 2x
d(area)/dx = 4x + 2
= 0 for a max/min
4x + 2 = 0
4x = -2
x = -1/2
also 2(2x + 2(x+1)) = 36
4x + 4x + 4 = 36
8x=32
x=4
Question makes no sense, check your typing or the wording of the question.
To minimize the sum of the areas of the two rectangles, we need to find the dimensions of each rectangle that will result in the smallest possible sum.
Let's assume the length of the first rectangle is x cm, then the width will be (x-1) cm, as given that each rectangle is 1 cm longer than its width.
The perimeter of a rectangle is given by: 2(length + width).
For the first rectangle, the perimeter can be expressed as: 2(x + (x-1)) = 2(2x-1) = 4x - 2.
Similarly, the length of the second rectangle will be (36 - x), and the width will be ((36 - x)-1) = (35 - x).
For the second rectangle, the perimeter can be expressed as: 2((36 - x) + (35 - x)) = 2(71 -2x) = 142 - 4x.
Since the total wire length is 36 meters, it can also be expressed as 3600 cm.
So, the sum of the two perimeters should be equal to the total wire length: (4x-2) + (142-4x) = 3600.
Simplifying the equation: 4x - 2 + 142 - 4x = 3600.
Combining like terms, we get: 140 = 3600.
Since this equation is not valid, it means that we have made a wrong assumption about the length of the first rectangle (x). To find the correct length, we need to consider the fact that the sum of the perimeters should be equal to the wire length.
So, let's assume the length of the first rectangle as y cm, then the width will be (y-1) cm.
The perimeter of the first rectangle will be: 2(y + (y-1)) = 2(2y-1) = 4y - 2.
The length of the second rectangle will be (36 - y), and the width will be ((36 - y)-1) = (35 - y).
The perimeter of the second rectangle will be: 2((36 - y) + (35 - y)) = 2(71 -2y) = 142 - 4y.
Setting up the equation based on the sum of perimeters: (4y-2) + (142-4y) = 3600.
Simplifying the equation: 140 = 3600.
Again, this equation is not valid, which means our assumption for the length of the second rectangle is also incorrect.
Let's evaluate another approach:
Let's assume the length of the first rectangle as a cm, then the width will be (a-1) cm.
The perimeter of the first rectangle will be: 2(a + (a-1)) = 2(2a-1) = 4a - 2.
The length of the second rectangle will be (36 - a), and the width will be ((36 - a)-1) = (35 - a).
The perimeter of the second rectangle will be: 2((36 - a) + (35 - a)) = 2(71 -2a) = 142 - 4a.
Setting up the equation based on the sum of perimeters: (4a-2) + (142-4a) = 3600.
Simplifying the equation: 140 = 3600.
Again, this equation is not valid.
It seems that we are not getting a valid equation that would give us the values for the lengths of the rectangles to minimize the sum of their areas.
It is possible that the given information or problem statement might have errors. Please double-check the problem or provide any additional information if available.
To minimize the sum of the areas of the two rectangles, we need to find the lengths of the two pieces of wire that will maximize the area of each rectangle.
Let's say the length of the first piece of wire is x meters. Then, the length of the second piece of wire will be 36 - x meters.
Each piece of wire is bent to form a rectangle, and we are given that each rectangle is 1 cm longer than its width. So, the dimensions of the first rectangle would be x/2 meters by (x/2 + 0.01) meters, and the dimensions of the second rectangle would be (36 - x)/2 meters by ((36 - x)/2 + 0.01) meters.
The formula for the area of a rectangle is length multiplied by width. Therefore, the area of the first rectangle would be (x/2) * (x/2 + 0.01) square meters, and the area of the second rectangle would be ((36 - x)/2) * ((36 - x)/2 + 0.01) square meters.
To minimize the sum of the areas, we need to find the values of x that will minimize the individual areas. To do that, we can take the derivative of each area expression with respect to x and set it equal to zero to find the critical points.
For the first rectangle area:
A1 = (x/2) * (x/2 + 0.01)
dA1/dx = 0.01 - x/2 = 0
Solving for x:
0.01 = x/2
x = 0.02
For the second rectangle area:
A2 = ((36 - x)/2) * ((36 - x)/2 + 0.01)
dA2/dx = 0.01 - (36 - x)/2 = 0
Solving for x:
0.01 = (36 - x)/2
x = 35.99
Since the length of the wire cannot be 35.99 meters, we round it to 36 meters.
Therefore, the two pieces should be 0.02 meters and 36 - 0.02 meters.