A rancher has 1,800 linear feet of fencing and wants to enclose a rectangular field and then divide it into two equal pastures with one internal fence parallel to one of the rectangular sides. what is the maximum area of each pasture? 5400

Answer: 67.500 ft²

Explanation:

1) Name the dimensions using variables:

y: length of the rectangular field

x: widht of the rectangular field

2) Model the amount of fence used by the two equal pastures:

two sides and one internal fence: 2x + x = 3x

two lengths: 2y

⇒ 3x + 2y = 1800 ← linear feet of fence

y = 1800 / 2 - 3x/2 ← solving for y

y = 900 - 3x/2

3) Area of each pasture

A = x (y/2) ← half ot xy

A = x (900 - 3x/2) ← replacing y with 900 - 3x/2

A = 900x - 3x² / 2 ← using distributive property

4) Maximum area ⇒ A' = 0

A' = 900 - 3x ← derivative of the polynomial 900x - 3x² / 2

900 - 3x = 0

⇒ 3x = 900

⇒ x = 900/3

⇒ x = 300

4) Determine y

y = 900 - 3x / 2 = 900 - 3 (300) / 2 = 900 - 450 = 450

5) Area of each pasture

A = xy/2 = 300 * 450 / 2 = 67500 ← final answer