Max Area of enclosed area, given fence down middle

[confused_07]

A farmer has 49=80 meters of fencing. He wishes to enclose a rectangular plot of land and to divide the plotinto three equal rectangles with two parallel lengths of fence down the middle. What dimensions will maximize the enclosed area? Besuer to verify that you have found the maximum enclosed area.

So far I got:

A= (xy)/3 (to give me the three rectangles)
P=2x+4y (x= top and bottom, 4y= 4 parallel fences)

Solve for y= (1/4)(480-2x) = (1/2)(240-X)

Plug into Area:

A= {[x * (1/2)(240-x)] / 3}
= {[(1/2)(240x-x^2)] / 3}

Am I right so far? Or am I wrong on my calculations?

 

[soroban]

Re: Max Area

Hello, confused_07!

A farmer has 480 meters of fencing.
He wishes to enclose a rectangular plot of land
and to divide the plotinto three equal rectangles
with two parallel lengths of fence down the middle.
What dimensions will maximize the enclosed area?
Be sure to verify that you have found the maximum enclosed area.

Your work is correct . . .

But I would let: \(\displaystyle \:A \:=\:xy\) . . . the total area.

Then the fencing gives us: \(\displaystyle \:2x\,+\,4y\:=\:480\;\;\Rightarrow\;\;y\:=\;120\,-\,\frac{1}{2}x^2\)

Substitute: \(\displaystyle \:A\;=\;x\left(120\,-\,\frac{1}{2}x\right) \;=\;120x\,-\,\frac{1}{2}x^2\)

. . . and so on . . .