If 500 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 two decimal places.
Ah, the classic paradox of trying to maximize something while being fenced in! So, if we have a rectangular plot of land with one side being the river, we need to find the dimensions that give us the maximum area. Let's solve this puzzle.
Let's assume that the length of the plot is x feet and the width (parallel to the river) is y feet. Since we're encircling the plot with fencing, we have two equal lengths and two equal widths, totaling four sides. This means that 2x + y = 500 (because the perimeter is given as 500 feet).
Solving for y, we get y = 500 - 2x.
Now, the area of the plot, A, can be calculated as A = x * y = x * (500 - 2x).
To maximize the area, we need to find the value of x that gives us the maximum value of A. So, let's differentiate A with respect to x and set it equal to zero to find the critical point:
dA/dx = 500 - 4x.
Setting dA/dx = 0, we find 500 - 4x = 0. Solving for x, we get x = 125.
Plugging x = 125 into our equation for y, we get y = 500 - 2x = 500 - 2 * 125 = 250.
Therefore, the maximum area that can be enclosed is A = x * y = 125 * 250 = 31,250 square feet.
So, the maximum area that can be enclosed is 31,250 square feet. Hope this answer didn't "fence" you in with too much information!
To find the maximum area that can be enclosed with 500 feet of fencing, we need to find the dimensions of the rectangular plot that will give us the largest possible area.
Let's assume the length of the rectangular plot is L, and the width is W.
Since the plot borders a river, we only need to enclose three sides with fencing - both widths and one length. Therefore, the perimeter is given by:
Perimeter = 2L + W = 500 feet
We can rearrange the equation to solve for one of the variables. Let's solve for W:
W = 500 - 2L
To maximize the area, we can use the fact that the area of a rectangle is given by:
Area = Length x Width
Substituting the expression for W, we get:
Area = L x (500 - 2L) = 500L - 2L^2
To find the maximum area, we need to find the vertex of the parabola formed by this equation. We can do this by finding the value of L that maximizes the area using the formula:
L = -b/2a
In our case, a = -2 and b = 500. Substituting these values into the formula, we get:
L = -500 / (2 * -2) = 125
So, the length that maximizes the area is 125 feet.
To find the corresponding width, we can substitute this value back into the equation for W:
W = 500 - 2L = 500 - 2(125) = 250
Therefore, the width is 250 feet.
Finally, we can calculate the maximum area by multiplying the length and width:
Area = L x W = 125 * 250 = 31,250 square feet.
Hence, the maximum area that can be enclosed with 500 feet of fencing is 31,250 square feet.
To find the maximum area that can be enclosed with 500 feet of fencing, we need to set up an equation based on the given information.
Let's assume that the length of the plot of land parallel to the river is L, and the width of the plot of land perpendicular to the river is W.
The perimeter of the plot of land is the sum of all the sides. In this case, it would be:
2L (for the length parallel to the river) + W (for the width perpendicular to the river) = 500
Simplifying the equation, we have:
2L + W = 500
Now, we need to express the area of the plot of land in terms of L and W. The area of a rectangle is given by the product of its length and width, so the area, A, is:
A = L * W
To find the maximum area, we need to express one variable (either L or W) in terms of the other variable using the perimeter equation and substitute it into the area equation.
Solving the perimeter equation for L, we have:
2L = 500 - W
Dividing both sides by 2:
L = (500 - W)/2
Now, substitute this expression for L in the area equation:
A = ((500 - W)/2) * W
To find the maximum area, we will differentiate the area equation with respect to W and find its critical points.
dA/dW = (500 - W)/2 - W/2
Setting dA/dW equal to zero and solving for W:
(500 - W)/2 - W/2 = 0
500 - W - W = 0
500 - 2W = 0
2W = 500
W = 250
Now substitute the value of W back into the perimeter equation to find L:
2L + 250 = 500
2L = 500 - 250
2L = 250
L = 125
So, the maximum area that can be enclosed with 500 feet of fencing is when the width, W, is 250 feet and the length, L, is 125 feet.
Substituting the values into the area equation:
A = (125 * 250) = 31250 square feet
Therefore, the maximum area that can be enclosed is 31250 square feet.
500 = L + 2 w
so
L = 500 - 2 w
Area = L w = (500 - 2 w)w = 500 w - 2 w^2
where is vertex of that parabola?
2 w^2 - 500 w = - A
w^2 - 250 w = -(1/2) A
w^2 - 250 w + 125^2 = -(1/2) A + 15625
(w-125)^2 = -(1/2)( A - 31250)
so
A = 31250 at vertx