Answer:
50 ft × 75 ft
Step-by-step explanation:
Let's say the width of each pen is W, and the length of each pen is L. The total length of the enclosure is therefore 2L, so its area is:
A = W (2L)
A = 2WL
Four of the walls will have a dimension of L. Three of the walls will have a dimension of W (one wall on each end and the wall between the pens). The amount of fencing is:
300 = 3W + 4L
Solve for either variable and substitute into the area equation. If we solve for L:
300 − 3W = 4L
L = 75 − ¾ W
Substituting:
A = 2W (75 − ¾ W)
A = 150W − 1.5 W²
We want to maximize the area. There are two ways we can do this. The first is to notice that this equation is a downward parabola, and the maximum of a downward parabola is at the vertex.
The vertex of y = ax² + bx + c is at x = -b/(2a).
W = -150 / (2 × -1.5)
W = 50
The other way is to use calculus. Take the derivative and set to 0:
dA/dW = 150 − 3W
0 = 150 − 3W
W = 50
L = 75 − ¾ W
L = 37.5
The width of each pen is 50 ft, and the length of each pen is 37.5 ft. Which means the length of the entire enclosure is 75 ft.
