Question

A rancher has 120 feet of fence with which to enclose three sides of a rectangular pasture (the fourth side is a cliff wall and will not require fencing). Find the dimensions of the pasture with the largest possible area. (For the purpose of this problem, the width will be the smaller dimension (needing two sides); the length with be the longer dimension (needing one side).)

length = _____ feet
width = ______feet

What is the largest area possible for this pasture?
area =______ feet-squared

Enter your answers as numbers. If necessary, round to the nearest hundredths.

Answer 1

2w + L = 120

L = 120-2w

Area(A) = L x W

substitute for l in area equation

A = (120-2w) w

A = 120w -2w^2

Take the first derivative of A

A' = 120 -4w

at A' = 0 and solve for w

4w = 120

w = 30

L = 120 - 60 = 60

length = 60 feet

width = 30 feet

area = 30 x 60 = 1800 square feet

Answer 2

Answer:

Step-by-step explanation:

Let L represent the length of the rectangular fence.

Let W represent the width of the rectangular fence.

The perimeter of the fence would be 120 feet.

Since the length would be one side and the width would be 2 sides, the sum of the 3 sides would be

L + 2W = 120

L = 120 - 2W

Area = length × width

Area = W(120 - 2W)

Area = 120W - 2W²

For maximum area, dA/dW = 0

Differentiating, it becomes

120 - 4W = 0

4W = 120

W = 120/4

W = 30 feet

L = 120 - 2W = 120 - 30 × 2

L = 120 - 60 = 60 feet

The maximum largest area possible for this pasture is

Area = (120 × 30) - 2(30)²

Area = 3600 - 1800 = 1800 square feet