find the max area of a fenced in area along a riverbank with only 500m of fencing?

Question

use no fencing along the riverbank

Answer 1

There are many possible shapes that you can use a fence. Its may take sometime to prove it by I can tell you that the best shape to maximize the area is a semi-circle. Thus all the above answers (except that one right above this) are wrong as you shall see for yourself.

ok... so we have a semi circle with a circumferences of 500. The Circumferences of a semi(half) circle = (2πr)/2 = πr

πr = 500 >> Radius of semi circle = r= 500/π

So now we have the radius, we can find the area.

Area of semi circle = (πr^2)/2 = (π/2)(500/π)^2 = 125000/π
= 39788.74 m^2

Obviously you can see that the Above Area is greater than all the above answers and indeed the maximum area possible.

-------------------------------------------------------------------------
The proof if you are really interested has something to do with the following: The Area created by the fence can be a rectangle as some people suggested. But how about a polygon? Will a 5 sided polygon (pentagon) maximize the area. Then you will realize there is a trend that as the number of sides of the sides increase, the area increases given a constant circumference. Therefore, an infinite number of sides, that is a circle will give us the maximum area. In your case is a semi-circle, since you want it along a riverbank.

Answer 2

First, we draw a diagram; a 3 sided fence (since the 4th side doesn't exist due to being along a riverbank). Visualize the fenced area as having a length (L) and a width (W). Two of these lengths and one of these widths have to add up to 500.

That is, since there is 500m of fencing

2L + W = 500

The area of the fenced area, given the variables we introduced is:
A = LW

This is what we want to maximize. First, we have to solve for one variable in terms of the other. Since

2L + W = 500, then W = 500 - 2L

Since A = LW, then
A = L(500 - 2L)
A = 500L - 2L^2
So now, this is our function, and we change A to A(L) since it's now consisting of a single variable.

A(L) = 500L - 2L^2

To find the max, take the derivative and make it 0.
A'(L) = 500 - 4L
0 = 500 - 4L
4L = 500
L = 500/4 = 125

So the maximum occurs at L = 125. Since we want the actual max area and not the measurements, we just plug in L = 125 for our created function.

A(125) = 500(125) - 2(125)^2

And then you just solve it for there, remembering to express your answer as "meters squared".

Answer 3

So you have only three sides. two of the sides are of x length, and the third side has a length of 500-2x. The formula for area gives you the following:
y = x(500-2x)
y = 500x-2x^2 = -2x^2 + 500x

This is a parabola which opens down. The positive y-values of the points on the parabola are all of the possible areas of the fenced in area. The maximum value of the area is the y-coordinate of the vertex of the parabola. Following is the procedure for finding this maximum value:

x-coordinate of vertex = -b/2a = -500/2 (-2) = 125
y-coordinate of vertex = -2(125)^2 + 500(125) = 31,250 sq meters - this is the maximum area

Answer 4

The other answerers are assuming that you have to have a rectangular shaped area. While this is a common problem in textbooks and probably in the original problem, it is not a requirement in your statement of the problem. With that in mind, we can have an unknown shape as well.

It turns out that the shape in the situation you describe with maximal area given a fixed fence length is a half-circle. So we just find the radius = 2*500/(2*pi) = 500/pi, and the area is (pi/2)*(500/pi)^2 = 125000/pi square meters, which is certainly bigger than the 31250 square meters an optimal rectangle gives you. So there! Muahahahaha!!!

Answer 5

Let
x - width
y - length
w - area

2x + y = 500 => y = 500 -2x
w = xy = x(500-2x) = 500x - 2x^2
dw/dx = 500 - 4x = 0 => x = 500/4 = 125

y = 500 - 2 x 125 = 250

w = 125 x 250 = 31250 m^2

Answer 6

You will acheive an area of 31,250 m squared if you make the front fence 250m and each side 125m. This yields the most area.