I want a rectanglular pen with area = 50 square feet and to save on fencing I will build it next to the garage. How should I design it so I use the least amount of fence-only 3 sides will be needed next to the garage. How can I figure out that I have found the maximum area for the pen?
Thanks much!
lola
assuming the garage will be one of the long sides (otherwise lots of extra fencing will be used), then if the long side is x, the short side will be 50/x.
So, the amount of fence, f, is
f(x) = x + 2(50/x) = x + 100/x
df/dx = 1 - 100/x^2
we want minimum fence, so df/dx = 0:
x = 10
So, the pen is 10x5, using 20 ft of fence.
Note that the square is the figure of maximum area using a given perimeter. Or, conversely, a square haas the minimum perimeter to enclose a given area.
Your pen is just two squares, where each encloses half the area. The garage wall helps out here.
To find the maximum area for the pen, given that you have to use only 3 sides of the fence next to the garage, you need to determine the dimensions of the rectangle that will give you the largest possible area.
Let's assume the length of the rectangle is "L" and the width is "W". Since the area of a rectangle is given by the product of its length and width, we have L * W = 50.
To minimize the amount of fence used, you need to account for the 3 sides adjacent to the garage, which means only one side will require fencing. Since a rectangle has opposite sides of equal length, you can choose one side (let's say the width, W) to be the side requiring a fence.
Given this information, we can express the perimeter of the rectangle using the length (L) and the width (W) as: Perimeter = L + 2W.
Considering that you want to minimize the fence usage, it means you want to minimize the perimeter. Since the length and width are related by the equation L * W = 50, we can substitute L in terms of W as L = 50/W.
The perimeter equation now becomes: Perimeter = 50/W + 2W.
To find the minimum perimeter, we can take the derivative of the perimeter equation with respect to W, set it to zero, and solve for W. This will give us the width for which the perimeter is minimized.
Taking the derivative of the perimeter equation with respect to W: d(Perimeter)/dW = -50/W^2 + 2.
Setting the derivative equal to zero: -50/W^2 + 2 = 0.
Simplifying: -50/W^2 = -2.
Multiplying both sides by W^2: -50 = -2W^2.
Dividing by -2: 25 = W^2.
Taking the square root of both sides: W = ±5.
Since the width of a rectangle cannot be negative, we discard the negative solution, leaving us with W = 5.
Now that we have the width, we can find the length by substituting W = 5 into the equation L * W = 50: L * 5 = 50.
Simplifying: L = 10.
Therefore, the dimensions of the rectangle that will give you the maximum area while using the least amount of fence are length (L) = 10 feet and width (W) = 5 feet.
To verify that this gives you the maximum area, calculate the area using these dimensions: Area = L * W = 10 * 5 = 50 square feet.
Hence, you have found the maximum area for the pen, which is 50 square feet, while using the least amount of fence.