A rancher wishes to enclose a rectangular partitioned corral with 1,872 feet of fencing. (See the illustration.) What dimensions of the corral would enclose the largest possible area?

Answer 1

No illustration is shown, but assuming one partition, then

call the total fencing as 2L+3W with one W as the partition distance.

2L+3W = 1872

L= (1872-3W)/2

Area = LW = [(1872-3W)/2](W) = 1872W/2 -3W^2/2

A= 936W -3W^2/2

take the derivative and set = 0

A' = 936-3W = 0

W= 936/3 = 312 feet Wide

L=(1872-3(312))/2 = 936-468 = 468 feet Long

A= LW = 468(312) = 146,016 ft^2= max Area

if there are more partitions such as 2, then recalculate using 2L+4W = 1872

if there are n partitions, then use 2L+(2+n)W=1872

312 = (2/3)468

if there are no partitions, a square gives the largest area, with each side = 1872/4 =468 feet

0 partition, one side is 2/2=1 = same as the perpendicular side

1 partition, one side is 2/3 of the other perpendicular side

2 partitions one side is 2/4 or 1/2 the other side

3 partitions, one side is 2/5 the other side

n partitions, one side is 2/(2+n) of the other side

6 partitions one side is 2/8 = 1/4 the other side

2L + 8W = 1872

L = (1872-8W)/2 = 936-4W

A= (936-4W)W = 936W-4W^2

A'=936 -8W=0

W=936/8=117

L=(1872-8(117))/2 =936-468=468

W=(1/4)468=117

those are the solutions if the partitions run only in one direction

if they criss cross, rework the problem differently