A rancher has 280 feet of fence with which to enclose three sides of a rectangular field (the fourth side is a cliff wall and will not require fencing). Find the dimensions of the field 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 field?

area = feet-squared

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

Question

Mathematics

Efrain Hawkins

22 April, 22:57

A rancher has 280 feet of fence with which to enclose three sides of a rectangular field (the fourth side is a cliff wall and will not require fencing). Find the dimensions of the field 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 field?

area = feet-squared

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

+5

Answers (1)

Know the Answer?

Yazmin Davenport · Answer 1

x = 140 ft

w = 70 ft

A (max) = 9800 ft²

Step-by-step explanation:

We have:

280 feet of fence to enclose three sides of a rectangular area

perimeter of the rectangle (3 sides) is

p = L = x + 2w w = (L - x) / 2 w = (280 - x) / 2

where:

x is the longer side

w is the width

A (x, w) = x*w ⇒ A (x) = x * (280 - x) / 2 ⇒ A (x) = (280x - x²) / 2

Taking derivatives on bth sides of the equation

A' (x) = (280 - 2x) * 2 / 4 A' (x) = 0 (280 - 2x) = 0

280 - 2x = 0 x = 280/2

x = 140 ft

And w = (280 - x) / 2 ⇒ w = (280 - 140) / 2

w = 70 ft

A (max) = 9800 ft²