# Find the dimensions of the field that maximizes the total area

• Oct 2nd 2012, 06:06 AM
spyder12
Find the dimensions of the field that maximizes the total area
A rectangular field is to be subdivided in 2 equal fields. There is 8350 feet of fencing available.
Find the dimensions of the field that maximizes the total area. (List the longer side first)
Width =_____________ feet
Length =________________ feet
What is the maximum area ? Area =_________________
• Oct 2nd 2012, 06:49 AM
johnsomeone
Re: Find the dimensions of the field that maximizes the total area
width = w, length = l, P = 8350 (perimeter), A = total area. Units are everywhere feet.

P = 2w + 3l (That's 2w + 2l to bound the area, and another fence of length l down the middle to split the fields.)

Thus l = (P-2w)/3.

Note w>0, l>0. Thus also (P-2w)/3 >0, so w < P/2.

A = lw = w(P-2w)/3.

Translation: Maximize A(w) = w(P-2w)/3 where 0<w<P/2, and where P is constant.
• Oct 2nd 2012, 07:38 AM
richard1234
Re: Find the dimensions of the field that maximizes the total area
We have $P = 2w + 3l = 8350$, and $A = wl$.

By AM-GM inequality, $\frac{2w + 3l}{2} \ge \sqrt{6wl} \Rightarrow \sqrt{6wl} \le \frac{8350}{2}$

The equality case occurs when 2w = 3l.
• Oct 2nd 2012, 09:52 AM
Soroban
Re: Find the dimensions of the field that maximizes the total area
Hello, spyder12!

Did you make a sketch?

Quote:

A rectangular field is to be subdivided into two equal fields.
. .
Actually, the two subfields do not have to be equal!
There is 8350 feet of fencing available.

Find the dimensions of the field that maximizes the total area.
What is the maximum area?

Code:

      : - - - x - - - :       *-----*---------*       |    |        |     y|    |y        |y       |    |        |       *-----*---------*       : - - - x - - - :

The perimeter is: . $2x + 3y \:=\:8350 \quad\Rightarrow\quad y \:=\:\tfrac{8350-2x}{3}$

The area is: . $A \;=\;xy \;=\;x\left(\tfrac{8350-2x}{3}\right)$

. . . . . . . . . . $A \;=\;\tfrac{8350}{3}x - \tfrac{2}{3}x^2$

Maximize: . . $A' \;=\;\tfrac{8350}{3} - \tfrac{4}{3}x \;=\;0 \quad\Rightarrow\quad \tfrac{4}{3}x \:=\:\tfrac{8350}{3}$

. . . . . . . . . . $x \:=\:\frac{4175}{2} \:=\:2087\tfrac{1}{2}\text{ ft}$

. . . . . . . . . . $y \:=\:\frac{4175}{3} \:=\:1391\tfrac{2}{3}\text{ ft}$

$\text{Area } \;=\;xy \;=\;\left(\frac{4175}{2}\right)\left(\frac{4175}{ 3}\right) \;=\;\frac{17,\!430,\!625}{6} \;=\;2,\!905,\!104\tfrac{1}{6}\text{ ft}^2$