Trent wants to build a rectangular pen for his animals. One side of the pen will be against the barn; the other three sides will be enclosed with wire fencing. If Trent has 400 feet of fencing, what dimensions would maximize the area of the pen?
A square would maximize the area.
400/3 = ?
To maximize the area of the rectangular pen, we need to find the dimensions that will give us the largest possible area. Let's break down the problem step-by-step:
1. Let's assume the length of the pen as 'L' and the width as 'W'. We want to find the values of L and W that maximize the area.
2. We know that one side of the pen will be against the barn, and the other three sides will be enclosed with wire fencing. So, the perimeter of the pen will be the sum of the length of all four sides.
3. Since one side is against the barn, we can subtract the length of this side from the total perimeter. The remaining three sides, which need wire fencing, will have a combined length of 400 feet minus the length of the side against the barn.
4. The perimeter of the three sides with wire fencing is given by: Perimeter = L + 2W
5. We can set up the equation: L + 2W = 400 - L
6. Rearrange the equation to solve for L: L + L + 2W = 400
2L + 2W = 400
L + W = 200
7. Now we have the length in terms of the width, so we can substitute this expression for L in the area formula.
8. The area of the pen is given by: Area = length x width = L * W
9. Substitute L + W = 200 into Area = L * W: Area = (200 - W) * W
Area = 200W - W^2
10. The area equation is now in terms of a single variable, W. To find the value of W that maximizes the area, we can take the derivative of the area equation with respect to W and set it equal to zero. This will give us the critical points, which could be the maximum or minimum values.
11. Differentiating the area equation, we get: d(Area)/dW = 200 - 2W
12. Set d(Area)/dW = 0 and solve for W: 200 - 2W = 0
2W = 200
W = 100
13. Now that we have the value of W, we can substitute it back into the equation L + W = 200 to find L:
L + 100 = 200
L = 100
14. Therefore, the dimensions that will maximize the area of the pen are: Length = 100 feet, Width = 100 feet.
So, Trent should build a rectangular pen with dimensions 100 feet by 100 feet to maximize the area.