A rectangular piece of land is bordered on one side by a river. The other 3 sides are to be enclosed by 300 feet of fencing. What is the maximum area that can be enclosed?

Question

A rectangular piece of land is bordered on one side by a river. The other 3 sides are to be enclosed by 300 feet of fencing. What is the maximum area that can be enclosed?

Answer 1

Hello, Mads,

John has the correct answer (rather, I should say I got the same answer as he did, so I'm encouraged enough to include my own derivation). Perhaps my outline will help in some way.

Let L = Lenght, and W = Width. Area is L*W

The 300 feet of fencing must be 2W + 1L, since the second L is the river. Furthermore, the total is = 300 feet:

2W + L = 300

We can aslo say L = 300 - 2W, or even W = (300-L)/2 . One of these expressions will come in handy later. These equations set the constraint of how much fencing can be used for maximizing the area.

Area = L*W, and we want as large a value we can find with the limited fencing. Let's first eliminate one of the variables by substituting one of the expressions above. I'll choose to eliminate the W (Width).

Area = L*W

Area = L*(300-L)/2)

Area = (300L - L²)/2

Area = 150L - (1/2)L²

Written in standard form, we quikly see that this is a parabola:

Area = - (1/2)L² + 150L

We want to maximize the area, so let's take the first derivative and set it equal to 0, which will tell us where on this parabola we can find a zero slope. For a parabola, this will be the apex, the coordiantes of which will be the values of L and W that provide the greatest area possible.

Area' = -L + 150

0 = -L + 150

L = 150

The length of one side of the rectangle is 150 feet. that leaves 150 feet of fencing, so each edge (the Width) would be 75 feet.

The area is 150ft*75feet = 11250 ft², the maximum area taht can be made under the conditions provided.

Tey Desos graphing software and plot the area function, using x and y:

y = -(1/2)x²+ 150x

You'll see the parabala that tells us the area of the rectangle as a function of x, the Length. The apex is (150, 1125), [(Length, Area)].

Bob

Answer 2

Draw a sketch of the land / river. Let x be the vertical side extending from the river to the post, and let y be the side that is horizontal to the river. Because there is 300 feet of fence, 2x+y= 300 irrespective of the lengths x/y. The area function A = x*y. However, we want a function in terms of a single variable. Here it is in terms of two variables x and y. Note that if 2x+y=300, then y = 300-2x. Sub y into the area function to get an area function based on x alone. That is, Area(x) x*(300-2x) = 300x-2x².

Because a local max cannot occur unless the derivative of the function = 0, we inspect where that might be.

Differentiating the area function Area'(x) = 300 - 4x. Now set that equal to zero to determine where the local max or min might be. 300 - 4x = 0 implies x=75 feet. We can confirm it is a local max by seeing if the curve is concave down, or the second derivative is negative. Differentiating again, Area''(x) = -4. So the local max occurs when x = 75, making the y = 150 by algebraic solution.