Arturo wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three

Question

Maybe I'm just over thinking the question cause I really got tired of thinking about it. So please help

Answer 1

the side parallel to the barn will be twice the other sides and the cross section.

You said the "other three sides".... which in similar problems means you make two pens out of the rectanglular yard with a fence down the middle parallel to the other 2 sides

You can solve this as a calculus optimization problem

A = LW Fence = L+3W 200=L+3W L= 200-3W

reduce A to a function of just one variable

A = (200-3W)W = 200W -3W^2

take the derivative and set equal to zero and solve for W

200 -6W = 0

W = 200/6 = 33 1/3

L= 200-3(33 1/3) =200 - 100 = 100

maximum area = LW = 3,333 1/3 square feet

If you kept it though, as one big pen, then

A= (200-2W)W = 200W -2W^2

A' = 4W = 200, W=200/4 = 50

L = 200-2(50)= 100 feet

and max area = 5,000 sq. ft. for 200 feet of fence and a barn as one side of the rectangular pen

with sides 50 by 100 feet