2 lengths = 460 - 5 widths
1 length = 230- 2.5 width
Area = w(230 - 2.5w)
Area = -2.5w2 + 230w
Max at -b/2a where b = 230 and a = -2.5
-230/-5 = 46
Area = -2.5(46)^2 + 230*46 = 5290 ft^2
June K.
asked 11/13/23A javalina rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure below). There are 460 feet of fencing available to complete the job. What is the largest possible total area of the four pens?
2 lengths = 460 - 5 widths
1 length = 230- 2.5 width
Area = w(230 - 2.5w)
Area = -2.5w2 + 230w
Max at -b/2a where b = 230 and a = -2.5
-230/-5 = 46
Area = -2.5(46)^2 + 230*46 = 5290 ft^2
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.