A rectangular pig pen is made of 84 meters of fencing on three sides. The fourth side is a barn wall. Find the dimensions and greatest possible area of the enclosure.
2x+y = 84
a = xy = x(84-2x)
That's just a parabola. Find its vertex for maximum area.
To find the dimensions and greatest possible area of the enclosure, we need to consider the given information.
Let's start by visualizing the problem. We have a rectangular pig pen with three sides fenced and the fourth side being a barn wall.
First, let's assign variables to represent the dimensions of the rectangular pen. Let's call the length of the pen L and the width W.
We are given that the perimeter of the pen is 84 meters. Since the perimeter is the sum of all sides, we can write an equation based on this information:
2L + W = 84
Simplifying this equation by subtracting 2L from both sides, we get:
W = 84 - 2L
Now, we want to find the dimensions and greatest possible area of the enclosure. The area of a rectangle is given by the formula:
Area = Length * Width
Substituting the value of W from the previous equation into the area formula, we get:
Area = L * (84 - 2L)
Expanding this equation, we have:
Area = 84L - 2L^2
To find the greatest possible area, we can take the derivative of the area equation with respect to L and set it equal to zero:
d(Area)/dL = 0
Differentiating the equation, we get:
84 - 4L = 0
Solving for L, we find:
L = 21
Substituting this value back into the equation for W, we have:
W = 84 - 2(21) = 42
Therefore, the dimensions of the enclosure are L = 21 meters and W = 42 meters.
To find the greatest possible area, substitute the values of L and W into the area equation:
Area = 21 * 42 = 882 square meters
So, the greatest possible area of the enclosure is 882 square meters.