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# Fence problem?

ok, i need some help with this homework. i need to fence a portion of land so as to provide the greatest area. one side of the land boarders a river so it does not need to be fenced. i have 900 feet of fence to use, how do i do this? i remember that you get some formula and graph it and the vertex is the answer. help please?

### 5 Answers

- PuzzlingLv 71 decade agoFavorite Answer
Let L be the length of the fence parallel to the river and let W be the length of the two fences perpendicular to the river.

So 2W + L = 900

Solve for L:

L = 900 - 2W.

The area would be W x L.

Substituting in for L.

Area = W (900 - 2W)

= -2W² + 900W

Now maximize that by taking the derivative and setting it to zero.

-4W + 900 = 0

4W = 900

W = 225

If you don't know calculus, you could just graph y = -2x² + 900x, for various values of x. You'll see that it is an upside down parabola that is symmetric around the point x = 225 where it has its maximum.

However you figure it, the area will be maximized when you build a rectangle (half square) with dimensions of 225 feet x 450 feet. Total fencing will be 225 + 450 + 225 or 900 feet, and the area will be 101,250 sq. ft. (Notice how this is more than the intuitive answer of 300 x 300 or 90,000 sq. ft.)

- Anonymous1 decade ago
If you want to use a graph, you are probably looking for a rectangular area instead of a semi-circle. Let x, x and y be the sides of your area. The area is xy and the length of fence is 2x+y=900. You have two equations in two unknowns. y=900-2x, so the area is x(900-2x). You can graph that and find the maximum or if calculus is allowed, take the derivative and set it to zero.

- HelmutLv 71 decade ago
Assuming the river on one side is perfectly straight:

The maximum area will be enclosed by a semicircle of radius 900/PI, or 128,915.5 square feet.

- paulofhoustonLv 61 decade ago
Try 300'x300'x300' otherwise you will have to make a semicircle(pi*r^2) for the most land

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