A farmer with 1200ft of fencing wishes to enclose a rectangular area and then divide it into three pens with f?

A farmer with 1200ft of fencing wishes to enclose a rectangular area and then divide it into three pens with fencing parallel to one side of the rectangle. Determine the dimensions that will yield the largest possible total area.





think of it as a rectangle with three equal parts in it.|||the total fencing: 2a+3b=1200.


Let x=b -%26gt;a = (1200-3x)/2


let f(x) is the area


f(x) = a*b (%26lt;- area of a rectangle)


f(x) = x * [ (1200-3x)/2]


find df(x)/dx = 600-3x. Set this equal 0. Find x


Since x = b, you can find a interm of b.


Dimensions are 200,and 300.|||Let x be the length of each pens be x, y be the length .


Perimeter =4x+y


4x+y=1,200


y=1,200-4x





Area = xy = x(1200-4x)=1,200x-4x^2


Area = 1200x - 4x^2


Differentiate with respect to x and set it equal to 0


dA/dx =1200-8x=0


8x=1200


x=150


y=1200-4(150)= 600





The dimension with largest area is 150 and 600 (parallel to one side)





d^2A/dx^2 =-8x = -8(150) %26lt; 0


This verifies that the area has been maximized.|||Let e = length of area, w = width of area, A = area of rectangle, f = total amount of fencing.





f = 2e+4w = 1200


e + 2w = 600


e = 600 - 2w





A = e*w = (600 - 2w)*w = 600w - 2w^2


Take the derivative of A.


A' = 600 - 4w


Set A' equal to 0 and solve for w.


600 - 4w = 0


600 = 4w


w = 150


A' is positive for w%26lt;150 and negative for w%26gt;150, so the maximum of A occurs at w = 150.





e = 600 - 2w = 600 - 2*150 = 300





The dimensions of the rectangle are 150 ft by 300 ft.|||Since you want to know the maximum TOTAL area, forget about the sub-dividing part (unless there is something missing from the problem statement). A square 300 ft on a side would give the maximum area. The total area would be 90000 ft^2.|||150x300


please dont listen to anyone else except ax. they're all wrong. please people, im in eighth grade and i did this in two minutes. it isnt that hard thank you ax for not being retarded.|||The Largest Possible area is yielded by a square (300x300x300x300). Therefore you would have 3 pens each 300x100.|||Do your own homework.