# Fence Calculus Optimization Problem Question: So, say you have about 200 feet of fence materials and you are wanting to construct three sides of a rectangular fence with a wall forming the fourth side, what is the maximum possible area/square footage of the fence?

Solution:

1. Determine the maximum area that your fence can be is to get a formula for the dependent variable A (area). For your fence, let x represent the two sides of fencing perpendicular to the wall and let y represent the side of fencing parallel to the wall. Since we know that the area of a rectangle is the product of its base and its height, the area of the rectangle is given by the formula A = xy.

2. Write A as a function of either x or y. In this case, we will use y. Since all 200 feet of fence are to be used, 2x + y =200, so y = 200-2x

3. Substitute this value of y into our formula for area A = xy to obtain A(x) = x(200 – 2x) = 200x – 2x^2

4. Determine the domain of the function A. What we are doing here is finding the maximum and minimum length that x (sides perpendicular to wall) can be, which is in this case, 0 to 100.

5. Compute the derivative of the function A(x) = (200x – 2x^2), which is = 200 – 4x.

6. Find any critical points by setting 200 – 4x equal to 0 and solving for x. In this case, there is only one critical point, 200 – 4x = 0 so x = 50.

7. Find the possible areas of of critical points. The extrema of A can occur only at x = 0, x = 50, or x = 100. You will now evaluate A at each of those points and determine the absolute maximum area that your fence can be. A(0) = 0(200)= 0 ft^2 A(50) = 50(100) = 5,000 ft^2 Absolute Maximum A (100) = 100(0) = 0 ft^2

Answer: A maximum area of 5,000 square feet can be obtained by making the sides fencing perpendicular to the wall, 50 feet long and by making the side parallel to the wall 100 feet long.

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