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Elenna [48]
2 years ago
10

A rancher has 1,800 linear feet of fencing and wants to enclose a rectangular field and then divide it into two equal pastures w

ith one internal fence parallel to one of the rectangular sides. what is the maximum area of each pasture? 5400
Mathematics
1 answer:
Lina20 [59]2 years ago
5 0
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
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