The owner of a ranch has 3000 yards of fencing material

[endoo]

The owner of a ranch has 3000 yards of fencing material to use to enclose a rectangular piece of grazing land along the straight portion of the river, and then subdivide this area by means of a fence running parallel to the sides. No fencing is require along the river.

What are the dimensions of the largest area that can be enclosed? What is the area?

 

[stapel]

So you've drawn the field enclosure, labelled the side with the river, drawn the center divider, picked variables for the length and width, labelled the fencing lengths, and... then what? What have you tried? How far have you gotten? Where are you stuck?

Please be specific. Thank you.

Eliz.

 

[endoo]

i think the length would be 1500, then the widths would be 500, 500, and 500. which adds up to 3000. but then to figure out the area would i just do length*width, or since there is that fence that is subdividing would the area equation be different

 

[stapel]

i think the length would be 1500, then the widths would be 500, 500, and 500....

Are you supposed to be using guess-n-check instead of calculus...?

Eliz.

 

[endoo]

umm no. i did:

perimeter=3w + l=3000
so l= 3000-3w

then i did area=2wl
A=(2w)(3000-3w)
=6000w-6w^2

derivative of that would be 6000-12w so w=500

am i on the right track or what? for the area would it be just 500*1500, or something different because of the fence subdividing.

 

[stapel]

It really does help to label things and make your reasoning clear.

I will guess that your picture looks something like the following:

-------------L----------
|       |              |
| w     | w          w |
|       |              |
~~~~~~~~~~~~~L~~~~~~~~~~
      r  i  v  e  r
~~~~~~~~~~~~~~~~~~~~~~~~

...where "w" stands for the width and "L" stands for the length. Then the "perimeter" (in quotes, because we really mean "fence length") is given by:

. . . . .3000 = L + 3w

Solving, we get:

. . . . .L = 3000 - 3w

The area A of any rectangular region is of course A = Lw. (I'm not sure where you're getting that you need to multiply the width by 2...?) Then the area equation is:

. . . . .A(w) = (w)(3000 - 3w) = 3000w - 3w²

Differentiate this, and solve for the min point (or just use algebra, and find the vertex).

Eliz.

 

[endoo]

k so i got length=1500yds and width=500yds

so area would then be: (1500)(500)=750,000 yds^2 correct?