Optimization Question? A farmer has 6000 m of fencing and wishes to create a rectangular filed subdivided into four congruent plots of land. Determine the dimensions of the each plot if the area to be enclosed is a maximum. A: 375 m by 600m
My Attempt:
Fencing:
6000 = 3x + 3y
(6000-3x)/3 = y
Area 
A = xy
A = x(6000-3x)/3
A' = (18,000 - 18x)/9
0 = 18,000 - 18x
x = 1000
why do i keep getting this wrong? Thank you in advance!
A: A rectangular region may be divided into four congruent pieces in many ways.  Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized.  (A and 4A are simultaneously maximized.)
$2 \times 2$: Let $x,y$ be the lengths of the enclosed region.  The enclosed area is $A = xy$.  The fencing for the perimeter and the internal fences is $P=3x+3y=6000 \text{ m}$.  We see that $y = 2000 \text{ m} -x$ so $A = x(2000 \text{ m} - x)$ which is a parabola with maximum at $x=1000 \text{ m}$ so $y=1000 \text{ m}$.  Each plot is then $500 \text{ m} \times 500 \text{ m}$
$1 \times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots.  Then again $A = xy$ and now $P = 5x+2y = 6000 \text{ m}$.  This time, $y = 3000 \text{ m} - (5/2)x$, so $A = x(3000 \text{ m} - (5/2)x)$.  This parabola takes a maximum when $x=600 \text{ m}$ so $y=1500 \text{ m}$.  Each plot is then $600 \text{ m} \times 375 \text{ m}$
Your error is that you're subdividing incorrectly.
Edit:  Divided $y$ among the lengths of each of the four subparcels.
Edit:  Corrected a stupid units transposition.
