area of a rectangular region: a farmer wishes to create two rectangular regions bordering a river, by three fences perpendicular to the river and one connecting them. suppose that x represents the length of each of the three parallel pieces of fencing. she has 600 feet of fencing available.
a) what is the length of the remaining piece of fencing in terms of x?
b) determine a function A that represents the total area of the enclosed region.
c) give any restrictions on x
d) what dimensions of the total enclosed region would give an area of 22,500 feet squared?
e) what is the maximum area that can be enclosed?
600 feet total fence
3 sides of x feet
remaining side: 600 - 3x
a(x) = x(600-3x) = 3x(200-x)
naturally, 3x < 600, so x < 200 assuming an infinitesimally thin fence and poles of zero diameter. :-)
22500 = 3x(200-x)
-3x^2 + 600x - 22500 = 0
x = 50 or 150
max area achieved at x = 100
a(100) = 30,000
a) The length of the remaining piece of fencing can be found by subtracting the lengths of the three parallel pieces of fencing from the total available fencing length of 600 feet. Since each of the three parallel pieces has a length of x, the remaining piece of fencing length is 600 - 3x.
b) To determine a function A that represents the total area of the enclosed region, we need to find the dimensions of the rectangular regions. Let's assume the width of each rectangular region is y. The length of the total enclosed region can be represented by 2y + x. Therefore, the function A can be expressed as A = (2y + x) * y.
c) There are a few restrictions on x. Firstly, since the farmer wants to create two rectangular regions, the value of x should be less than half of the total available fencing length. So, we can write the inequality: x < 300. Secondly, since the remaining piece of fencing is not mentioned to be of a different length, we can assume it is also x. So, x should be less than or equal to 200 feet to allow for the construction of a rectangle.
d) To find the dimensions of the total enclosed region that would give an area of 22,500 feet squared, we can substitute the given area into the area function A: (2y + x) * y = 22,500. By rearranging the equation and solving for y in terms of x, we get y = (22,500 / (2x + x)) = (22,500 / 3x). This gives us the relationship between y and x.
e) To find the maximum area that can be enclosed, we need to optimize the area function A. Since A = (2y + x) * y, we can use calculus to find the maximum. Taking the derivative of A with respect to y and setting it equal to zero, we can find the critical points. Then, we can substitute these critical points back into the area function to find the maximum area. However, since this is a step-by-step bot, I can provide the relationship between y and x, but I cannot calculate the maximum area without additional information or using calculus.
a) To find the length of the remaining piece of fencing, we need to subtract the length of the three parallel pieces of fencing from the total available fencing. Each of the three parallel pieces of fencing has a length of x, so the remaining piece of fencing would have a length of:
Remaining piece of fencing = Total available fencing - 3x = 600 - 3x
b) The total area of the enclosed region can be calculated by multiplying the length and width of the rectangular region. Let's denote the width of each rectangular region as y. So, the total area, A, can be expressed as:
A = 2xy + xy = 3xy
c) There are some restrictions on x and y. Since the farmer wants to create two rectangular regions, the width y should be positive. Also, the length of the remaining piece of fencing should be non-negative, so:
600 - 3x ≥ 0
3x ≤ 600
x ≤ 200
This means that x should be less than or equal to 200.
d) To find the dimensions of the total enclosed region that would give an area of 22,500 square feet, we can set the area equation equal to 22,500 and solve for x and y:
3xy = 22,500
xy = 7,500
We can choose any values for x and y that satisfy the above equation. For example, if we let x = 50, then y = 150 (or vice versa), the total enclosed region would have dimensions of 50 feet by 150 feet, resulting in an area of 22,500 square feet.
e) To find the maximum area that can be enclosed, we need to maximize the area function A = 3xy, while considering the restrictions on x. Since x is constrained to be less than or equal to 200, the maximum area can be achieved when x is at its maximum value, which is 200.
So, substituting x = 200 into the area equation:
A = 3xy = 3(200)y = 600y
To find the maximum area, we need to find the maximum value of y. However, there are no specific constraints provided for y, so it can be arbitrarily large. Therefore, the maximum area that can be enclosed is infinite.