# Thread: Need help understanding these types of problems: fencing 2 side-by-side areas

1. ## Need help understanding these types of problems: fencing 2 side-by-side areas

Farmer Ted wants to build two pastures side by side with shared wall between them. Ted has $4800 to spend. The fencing around the perimeter costs$3 per foot and the
shared fencing costs $2 per foot. What is the max area Farmer Ted can enclose? What are the dimensions of the pastures? I have some work already done that attempts to answer the problem 3x+2y=4800 -3(x+y=5) ----------- 2y=4800 -3y=-15 ---------- -y= 4785 Im pretty sure this is solvable using a quadratic equation because the notes i took in class are over a similar problem but its set up differently so the steps to solving it are different as well. Im not exactly sure what to do with the -y=4785 that i got or if its even set up right. 2. Originally Posted by Matts Farmer Ted wants to build two pastures side by side with shared wall between them. Ted has$4800 to spend. The fencing around the perimeter costs $3 per foot and the shared fencing costs$2 per foot. What is the max area Farmer Ted can enclose?
What are the dimensions of the pastures?

I have some work already done that attempts to answer the problem

3x+2y=4800
-3(x+y=5)
-----------
2y=4800
-3y=-15
----------
-y= 4785

Im pretty sure this is solvable using a quadratic equation because the notes i took in class are over a similar problem but its set up differently so the steps to solving it are different as well. Im not exactly sure what to do with the -y=4785 that i got or if its even set up right.
Let the total field be L (ft) * W (ft).

Let the length of the shared fence be W (ft).

Then

3 * 2 * (L + W) + 2 * W = 4800 → 3L + 4W = 2400 → W = 600 - 0.75*L .......................(1)

A = L * W ............. Using (1), we get

A = L * (600 - 0.75 * L) ...... This is a quadratic and continue.....

3. Originally Posted by Denis
Or since a square produces maximum area,
you can use side = L, thus perimeter 4L.

You can then forget about the shared fence:
instead, use 3 sides @ $3 and other side @$5.
You can't forget about the [cost of the] shared fence. The solutions don't involve pastures that are square-shaped.

That is, the pastures turn out not to be square-shaped.

4. Originally Posted by Denis
Or since a square produces maximum area,
you can use side = L, thus perimeter 4L.

You can then forget about the shared fence:
instead, use 3 sides @ $3 and other side @$5.
That logic will not work here because of the fixed (but variable) cost involved.

5. Originally Posted by Denis
Ahhh shucks......>|_
And to think you were outstanding in your field.

6. Originally Posted by Denis
Or since a square produces maximum area,
you can use side = L, thus perimeter 4L.

You can then forget about the shared fence:
instead, use 3 sides @ $3 and other side @$5.
Corner time mon ami.

I have no idea why a problem of optimization under constraint is posed in algebra, but constraints may, and usually do, affect the answer. Moreover the problem is badly worded because, as given, there is no indication what, if any, constraints are placed on the shape of either pasture. I don't know how to solve this problem without limitations on the geometry; it may require calculus of variations or some even more advanced technique.

If we set up two equal semicircles joined by a wall along the diameter, the cost equation is

$3 * \pi d + 2d = 4800 \implies d = \dfrac{4800}{3 \pi + 2} \implies$

the combined area is

$\pi \left ( \dfrac{2400}{3 \pi + 2} \right )^2 \approx 138,636\ sq.\ ft.$,

which considerably exceeds the area formed by rectangles. So I am quite confident the problem, at least as stated, is beyond the power of algebra to solve. I have an intuition, which I cannot prove, that the correct answer involves symmetric semi-ellipses divided along the major axis.

7. Originally Posted by JeffM
Corner time mon ami.

I have no idea why a problem of optimization under constraint is posed in algebra, but constraints may, and usually do, affect the answer. Moreover the problem is badly worded because, as given, there is no indication what, if any, constraints are placed on the shape of either pasture. I don't know how to solve this problem without limitations on the geometry; it may require calculus of variations or some even more advanced technique.

If we set up two adjacent semicircles of equal diameter divided by a diameter fence, the cost equation is

$3 * \pi d + 2d = 4800 \implies d = \dfrac{4800}{3 \pi + 2} \implies$

the combined area is

$\pi \left ( \dfrac{2400}{3 \pi + 2} \right )^2 \approx 138,636$,

which considerably exceeds the area formed by rectangles.
I concur. I saw the OP's attempt at the solution (3x + 2y = 1800) and inferred (never assumed) that a rectangle was "suggested" in class.

8. Also, there is this:

Originally Posted by Matts

Farmer Ted wants to build two pastures side by side with shared wall $\ \ \ \$ Here it is called a "wall." One would think there would be no fencing.

between them. Ted has $4800 to spend. The fencing around the perimeter costs$3 per foot and the

shared fencing $\ \ \ \$ But here it is called "shared fencing."

costs $2 per foot. . 9. Originally Posted by Denis Well, certainly looks like a rectangle was intended (as Subhotosh says); so 400 by 300 = 120,000 sq.ft. 800*$3 + 600*$3 + 300*$2 = \$4800

Corner??
138 > 120

10. Originally Posted by Denis
You've lost me Jeff: what meanest thou?
Farmer Ted is not too smart. He can get 238K sq ft with just a bit of cleverness. No one told him that sheep care whether their pastures are rectilinear. You are misleading farmer Ted, and he needs all the help he can get.

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