the length and width of the sheep pen ,which is adjusted so that the enclosed area is maximum?

Answer 1

I am assuming that your sheep pen must be rectangular, guessing from your use of length and width, and you are working with a set perimeter value. In this case, your pen should be a square of length and width each equal to 1/4 of your total perimeter. However, if you are free to maintain a pen of any shape, you get the most area per unit perimeter in a circle, of radius = 1/(2pi) times the perimeter.

Answer 2

a square (if you must have a length and width and not a circle) encloses the most area with a fixed perimeter

are you trying to get homework help?

Answer 3

Too easy for a given sheep pen length. Consider please this other question (similar, but containing a contraint):
" How can one shape for the maximum enclosed area a sheep pen 22m long, entouring a 5m side square box"
http://answers.yahoo.com/question/index?qid=20061122015003AAaG7aR&pa=FYd1D2bwHTHwIr5nHOM7QSUOKLD3WKx_rdkIEnZYQDBofQ--&paid=asked&msgr_status=

Answer 4

Given lengths of fencing, a square will provide the largest area.

;-D If you graph AxB, where A+B is constant, the largest product occurs when A=B. 1x99=99, 50x50=2,500.
49x51=2499.
;-D Other multiples are all less than 2,500.