1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ What are 375 m and 600 m ? They don't appear anywhere. $\endgroup$ – Claude Leibovici Jan 26 '14 at 8:14
  • $\begingroup$ If $ \ x \ $ and $ \ y \ $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2. $\endgroup$ – colormegone Jan 26 '14 at 8:18
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500? $\endgroup$ – Jessica Jan 26 '14 at 14:22
  • 1
    $\begingroup$ @Jessica: Draw the picture of the $1 \times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems... $\endgroup$ – Eric Towers Jan 27 '14 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.