Solve this problem algebraically, using your knowledge of derivatives. Do not solve by graphing.
If 800 feet of fencing is used to enclose a rectangular plot of land that borders a river, what is the maximum area that can be enclosed?
x = length , y = width
x + 2y = 800 ... x = 800 - 2y
area = x y = y (800 - 2y) = -2 y^2 + 800 y
setting the 1st derivative equal to zero will show the max value
-4 y + 800 = 0 ... y = 200
You have information about the perimeter and area. Use them both.
P=w + L + w
800 = 2w + L
800 - 2w = L
Area = L(w)
Does this twig any thoughts about what you have learned?
then you will need the first and second derivatives.
To find the maximum area that can be enclosed using 800 feet of fencing, we can use the concept of derivatives to optimize the area function.
Let's denote the width of the rectangular plot as 'w' and its length as 'l'. We know that the perimeter of the rectangular plot is 800 feet, which gives us the equation:
2w + l = 800 ----(1)
The area of the rectangle is given by the product of its length and width, so:
A = w * l ----(2)
To solve this problem algebraically, we need to express one variable in terms of the other. From equation (1), we have:
l = 800 - 2w
Substituting this expression for 'l' into equation (2), we obtain:
A = w * (800 - 2w)
A = 800w - 2w²
Now, let's find the derivative of the area function with respect to 'w'. Differentiating the function A = 800w - 2w², we have:
dA/dw = 800 - 4w
To find the maximum area, we set the derivative equal to zero:
800 - 4w = 0
Now solve for 'w':
4w = 800
w = 200
Substituting this value back into equation (1) gives us:
2(200) + l = 800
400 + l = 800
l = 400
Therefore, the dimensions of the rectangular plot that maximize the area are a width of 200 feet and a length of 400 feet. To find the maximum area, substitute these values into equation (2):
A = 200 * 400
A = 80,000 square feet
So, the maximum area that can be enclosed using 800 feet of fencing is 80,000 square feet.
To find the maximum area that can be enclosed with 800 feet of fencing, we should first express the problem as a function and then use derivatives to find its maximum value.
Let's call the width of the rectangular plot x and the length y. We know that the perimeter of the rectangular plot is 800 feet, which consists of two widths and two lengths:
2x + 2y = 800
Now we can isolate y in terms of x by rearranging the equation:
2y = 800 - 2x
Divide both sides by 2:
y = 400 - x
The area A of the rectangular plot is given by:
A = x * y
Substituting the expression for y, we have:
A(x) = x * (400 - x)
To find the maximum area, we need to find the critical points of this function. We can do that by taking the derivative of A(x) with respect to x:
A'(x) = 400 - 2x
Setting the derivative equal to zero and solving for x:
400 - 2x = 0
2x = 400
x = 200
Now, we need to check whether this critical point is a maximum or minimum by evaluating the second derivative of A(x). Taking the derivative of A'(x):
A''(x) = -2
Since the second derivative is negative, the critical point x = 200 corresponds to a maximum.
To find the corresponding value of y, we can substitute x = 200 into the previous expression for y:
y = 400 - x
y = 400 - 200
y = 200
Therefore, the maximum area that can be enclosed with 800 feet of fencing is obtained when the width is 200 feet, and the length is 200 feet. Substituting these values into the area formula:
A = x * y
A = 200 * 200
A = 40,000 square feet.