A pig rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to o?

A pig 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). He has 860 feet of fencing available to complete the job. What is the largest possible total area of the four pens?|||Hi,





x = width, which is used 5 times to make 4 pens.





(860 - 5x)/2 = length of pens





y = x(860 - 5x)/2





y = 430x - 5/2x虏





The maximum area occurs at the vertex , which is at x = -b/(2a) = -430/-5 = 86 feet width





(860- 5x)/2 - (860 - 5(86))/2 = 430/2 = 215 = combined width of all 5 pens





86 x 215 = 18,490 ft虏 %26lt;==ANSWER





I hope that helps!!! :-)|||There is no figure below.|||Is this from webwork? math115 lol :P