5W + 2L = 1200

2L = 1200-5W

L = 600 -5W/2

Area = LW = (600-5W/2)W = 600W -(5/2)W^2

A'(W) = 600-5W =0

5W =600

W = 600/5 = 120 m

L = 600-300= 300 m

max Area = 120(300) = 36,000 square meters

Jordan M.

asked • 01/20/21A farmer with 1200 m of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

Follow
•
1

Add comment

More

Report

5W + 2L = 1200

2L = 1200-5W

L = 600 -5W/2

Area = LW = (600-5W/2)W = 600W -(5/2)W^2

A'(W) = 600-5W =0

5W =600

W = 600/5 = 120 m

L = 600-300= 300 m

max Area = 120(300) = 36,000 square meters

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.