find the dimensions of the plot that will maximize...

[wind]

Hi, I don't really know what I'm doing for this problem, can soeone help me?

At an agricultural research facility an area of land (next to an existing fence) must be divided equally into 9 rectangular plots,fencing will surround each plot. if each plot must have an area of 36m², find the dimensions of the plot that will maximize the amount of fencing required.

Area of each plot

A=LW
36=LW
36/L=W

Perimeter
9w+13L=P
9(36/L) + 13L=P
324/L +13L=P

... :?

Pls Help!

 

[TchrWill]

Hi, I don't really know what I'm doing for this problem, can soeone help me?

At an agricultural research facility an area of land (next to an existing fence) must be divided equally into 9 rectangular plots,fencing will surround each plot. if each plot must have an area of 36m², find the dimensions of the plot that will maximize the amount of fencing required.

The area of each plot is x(y) = 36


With x = the dimension perpendicular to the given fence and y = the dimension parallel to the given fence, the total perimeter of added fencing becomes P = 12x + 9y.

On the assmption that you are seeking integer dimensions for the plots, the factors of 36 are 1, 2, 3, 4, 9, 18 and 36 resulting in perimeter (x,y) pairings of 1-36, 2-18, 3-12, 4-9, 6-6, 9-4, 12-3, 18-2 and 36-1 . Nine punches on the calculator will gve you your answer.

Maximum fencing derives from 36 x 1 plots (P = 441.

Minimum fencing derives from 6 6x 6 plots P = 126

 

[soroban]

Hello, wind!

At an agricultural research facility an area of land (next to an existing fence)
must be divided equally into 9 rectangular plots. .Fencing will surround each plot.
If each plot must have an area of 36m², find the dimensions of the plot that will
minimize the amount of fencing required.

I assume that the set-up looks like this . . .

. . ============================================= =============
      |     |     |     |     |     |     |     |     |     |
     y|    y|    y|    y|    y|     |y    |y    |y    |y    |y
      |     |     |     |     |     |     |     |     |     |
      *-----*-----*-----*-----*-----*-----*-----*-----*-----*
      : - - - - - - - - - - - -  x  - - - - - - - - - - - - :

There will be one length of fencing \(\displaystyle x\) m long
. . and ten lengths each \(\displaystyle y\) m long.

The total fencing is: \(\displaystyle \:F \:=\:x\,+\,10y\;\) [1]

The total area of the plots is: \(\displaystyle \,9\,\times\,36 \:=\:324\) m².
. . \(\displaystyle xy\:=\:324\;\;\Rightarrow\;\;y \:=\:\frac{324}{x}\;\) [2]

Substitute [2] into [1]: \(\displaystyle \:F \:=\:x\,+\,10\left(\frac{324}{x}\right)\:=\:x\,+\,3240x^{-1}\)

Differentiate and equate to zero: \(\displaystyle \:F' \:=\:1\,-\,3240x^{-2} \:=\:0\)

Multiply by \(\displaystyle x^2:\;\;x^2\,-\,3240\:=\:0\;\;\Rightarrow\;\;x^2\:=\:3240\;\;\Rightarrow\;\;\fbox{x\:=\:18\sqrt{10}}\)

Substitute into [2]: \(\displaystyle \:y \:=\:\frac{324}{18\sqrt{10}}\;\;\Rightarrow\;\;\fbox{y\:=\:\frac{9\sqrt{10}}{5}}\)

 

[wind]

Thanks TchrWill, soroban

actually the set up looks like this

I guess I should have drawn a pic


So then, like TchrWill said, would the total fencing be 9x and 12y?


F=9x + 12y
F=9x+ 12(324/x)
F=9x+ 3888/x
F=9x + 3888x^-1

F'= 9 - 3888/x²
0= 9 - 3888/x²
9x²= - 3888
x²= 432
x= root432

At this point I should use the second derivitive test to make sure that it is a min value right?

F''=7776/x³
F''=0.866

positive so local min

then substitute the x val in to F=9x + 12y

right? or does the pic not matter :?