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? Answer to the nearest square foot without commas. For example, if the answer is 1,000, write 1000.
p + 2 q = 800 so p = 800 - 2 q
A = p q = (800 - 2q)q
A = -2 q^2 + 800 q
now you could find the vertex of this parabola easily with Algebra 2 by completing the square but you said calculus so
dA/dq = 0 at max = -4 q + 800
or
q = 200
then
p = 800 - 400 = 400
so
A = 80,000 ft^2 or about two acres
80,000 ft^2 just like how we gonna go back to back
Well, if I were to enclose a rectangular plot of land that borders a river, I would make sure it's a pretty fancy plot. You know, like a VIP section for land. So, let's call it "Fancyland."
Now, in Fancyland, we have a total of 800 feet of fencing to work with. The fancy part is that one side of the plot is already taken care of by the river. So, we only need to fence the other three sides.
To maximize the area, we want to make the shape as close to a square as possible. Why? Because squares are super fancy. Plus, they have equal sides, which saves us some fencing.
So, let's assume that two sides of the plot have length "x" and the third side (the one opposite to the river) has length "y". Since we don't need to fence the river side, we know that "x" + "y" + "y" = 800 (the two "y" sides plus the "x" side).
Simplifying that, we get "x + 2y = 800". Now, we want to maximize the area, which is "A = xy". To solve this problem, we need to express "x" in terms of "y" or "y" in terms of "x".
Let's solve "x + 2y = 800" for "x". We get "x = 800 - 2y". Now we can substitute that into the area formula: "A = (800 - 2y)y".
To find the maximum, we could take the derivative and all that fancy math stuff, but instead, let's just use our intuition. We know that if we have a rectangular plot of land with sides "x" and "y", the area will be maximized when "x" = "y". So, let's set "x" = "y" and see what happens.
We get "A = (800 - 2y)y". Substituting "y" for "x" in the equation, we get "A = (800 - 2y)y = 800y - 2y^2".
Now, to find the maximum area, we need to find the value of "y" that makes "A" reach its highest point. We can use the graph of a parabola or complete the square, but let's use an easier approach. We just need to find the vertex of the parabola, and that will give us the maximum area.
The vertex occurs at "y = -b/2a" in the equation "A = ay^2 + by + c". In our case, "a" is -2, "b" is 800, and "c" is 0. Plugging those numbers in, we get "y = -(800)/(2*-2) = 200".
So, the value of "y" that maximizes the area is 200. Now we can substitute that back into the equation to find the maximum area: "A = (800 - 2(200))200 = 120,000 square feet."
Therefore, the maximum area that can be enclosed is 120,000 square feet. And let me tell you, with that much space, you can do some seriously fancy clown tricks in Fancyland!
To find the maximum area that can be enclosed, we need to determine the dimensions of the rectangular plot of land.
Let's assume the length of the plot of land is L and the width is W.
We know that the perimeter of a rectangle is given by the formula:
Perimeter = 2L + 2W
In this case, the perimeter is given as 800 feet, so we can set up the equation:
800 = 2L + 2W
Simplifying the equation:
2L + 2W = 800
L + W = 400
We want to maximize the area, which is given by the formula:
Area = L * W
To solve for the maximum area, we can rearrange the previous equation to solve for one variable in terms of the other, and substitute it back into the area formula.
Let's solve for L in terms of W:
L = 400 - W
Substituting this into the area formula:
Area = (400 - W) * W
We can now maximize the area by finding the vertex of the quadratic equation.
The vertex of a quadratic equation in the form of Ax^2 + Bx + C can be found using the formula:
x = -B / (2A)
In our case, A is -1, B is 400, and C is 0.
Plugging in these values:
W = -400 / (2 * -1)
W = 200
Now substitute this value back into the equation to find L:
L = 400 - 200
L = 200
So the maximum area that can be enclosed is given by:
Area = 200 * 200
Area = 40,000 square feet
Rounding the answer to the nearest square foot without commas gives us the final answer:
40000
To find the maximum area that can be enclosed by the fence, we need to consider the properties of a rectangle. Let's denote the width of the rectangle as "w" and the length as "l".
Since the border of the land is a river, we only need to fence three sides of the rectangle—the two widths and one length. The other length is already bordered by the river.
Given that 800 feet of fencing is used, we can set up the equation:
2w + l = 800
To find the maximum area, we need to express the area in terms of a single variable. In this case, we can express the length "l" in terms of the width "w":
l = 800 - 2w
Now, we can express the area "A" in terms of "w" using the formula: A = length × width
A = w * (800 - 2w)
To find the maximum area, we need to find the vertex of the quadratic function. We can do this by finding the x-coordinate of the vertex using the formula: x = -b / (2a), where a and b are the coefficients of the quadratic function.
In this case, the quadratic function is:
A = -2w^2 + 800w
Comparing this equation to the standard quadratic form ax^2 + bx + c, we find that a = -2 and b = 800.
Now, we can calculate the x-coordinate of the vertex:
w = -800 / (2*(-2))
w = -800 / (-4)
w = 200
Since "w" represents the width, we can substitute this value back into our equation for "l":
l = 800 - 2w
l = 800 - 2(200)
l = 800 - 400
l = 400
Therefore, the maximum area that can be enclosed is:
A = w * l
A = 200 * 400
A = 80,000 square feet
So, the maximum area that can be enclosed is 80000 square feet.