You have 800 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed? River X 800 - 2x The width, labeled x in the figure, is (Type an integer or decimal.) feet The length, labeled 800 - 2x in the figure, is feet. (Type an integer or decimal.) The largest area that can be enclosed is (Type an integer or decimal.) square
Let's call the width of the plot x feet.
Since the side along the river does not need to be fenced, this means that we only need to consider three sides of the plot - the two ends and the back. Therefore, the length of the plot will be 800 - 2x feet.
To find the area of the plot, we multiply the length by the width:
Area = (800 - 2x) * x
To maximize the area, we can take the derivative of the area function with respect to x and set it equal to zero:
d/dx [(800 - 2x) * x] = 800 - 4x = 0
Solving for x:
4x = 800
x = 200
So, the width of the plot is 200 feet.
The length of the plot is 800 - 2(200) = 400 feet.
The largest area that can be enclosed is 200 * 400 = 80,000 square feet.
To find the length and width of the plot that will maximize the area, we can use the formula for the area of a rectangle: A = length × width.
Let's assume the width of the plot is x feet. The length of the plot would then be 800 - 2x feet (since we are not fencing the side along the river).
Using the formula for the area, we can write the equation:
A = (800 - 2x) × x
To find the value of x that maximizes the area, we need to find the critical points of the function A(x).
Taking the derivative of A(x) with respect to x:
A'(x) = 800 - 4x
Setting the derivative equal to zero and solving for x:
800 - 4x = 0
4x = 800
x = 200
Now, we need to determine whether this critical point is a maximum, minimum, or neither. To do this, we can take the second derivative of A(x):
A''(x) = -4
Since the second derivative is negative, this means that x = 200 corresponds to a maximum area. Therefore, the width of the plot that will maximize the area is 200 feet, and the length of the plot would be 800 - 2x = 800 - 2(200) = 400 feet.
Now, we can calculate the largest area that can be enclosed by substituting the values of x and the length into the area formula:
A = (800 - 2x) × x
A = (800 - 2(200)) × 200
A = (800 - 400) × 200
A = 400 × 200
A = 80,000 square feet
So, the largest area that can be enclosed is 80,000 square feet.
To find the dimensions that will maximize the area of the rectangular plot, we can use the concept of derivatives from calculus. Let's assume the width of the plot is x feet.
Given that the length of the plot is 800 - 2x (since the side along the river doesn't need to be fenced), the area can be calculated by multiplying the length and width:
Area = x * (800 - 2x)
To find the value of x that maximizes this area, we need to find the critical points of the function.
First, let's express the area equation in terms of x and simplify it:
Area = 800x - 2x^2
To find the critical points, we need to take the derivative of the area equation with respect to x and set it equal to zero:
d(Area)/dx = 800 - 4x
Now, set the derivative equal to zero and solve for x:
800 - 4x = 0
4x = 800
x = 200
So, x = 200 feet is the width that maximizes the area. To find the length, substitute this value of x back into the length equation:
Length = 800 - 2x = 800 - 2(200) = 800 - 400 = 400 feet
Therefore, the width of the plot is 200 feet, the length is 400 feet, and the largest area that can be enclosed is:
Area = 200 * 400 = 80,000 square feet.