Maximizing area against a curved river

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Homework Help Overview

The problem involves a farmer who has two miles of fencing to enclose the maximum possible area for his cows, using three sides of the fence due to a river represented by the parabola y=4-x^2. The challenge is to determine the optimal configuration of the fence, with one end at the vertex of the parabola and the other end somewhere in the first quadrant.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to clarify whether the fence must be parallel to the axes and whether it should be positioned to the right of the parabola. There are mentions of using substitution and differentiating the area equation, while some express confusion about variable assignments and constraints imposed by the problem.

Discussion Status

The discussion is ongoing, with participants offering hints and nudges without providing direct solutions. Some guidance has been given regarding the setup of equations and the differentiation process, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants note that the problem is set in a calculus 1 class, which restricts the use of certain methods like Lagrange multipliers. This constraint influences the strategies being discussed.

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A farmer has two miles of fence, and wants to enclose the max possible area for his cows. He will use fence on only three sides because the fourth side is a river that flows through his land. The path of the river is the parabola y=4-x^2. One of the ends of the fence is on the vertex of the parabola. and the other end of the fence is on the parabola somewhere in the first quadrant. Find the exact value of the max area.

Here an image of how i labeled the variables on the graph and the two equations I got:
http://img139.imageshack.us/img139/1655/problemnl2.png

2 = t + 2s - a

The professor only told us to label that point (s,t) and that we'd have to differentiate the integral eventually.

I'm confused about which letters should be treated as variables because I know I have to work with the perimeter equation to substitute it in. Any helpful hints or nudges in a good direction?
 
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Welcome to PF!

Mylala88 said:
A farmer has two miles of fence, and wants to enclose the max possible area for his cows. He will use fence on only three sides because the fourth side is a river that flows through his land. The path of the river is the parabola y=4-x^2. One of the ends of the fence is on the vertex of the parabola. and the other end of the fence is on the parabola somewhere in the first quadrant. Find the exact value of the max area.

Hi Mylala88! Welcome to PF! :smile:

I'm a little confused … could you please clarify:

does the problem specify that the fence must go parallel to the axes (because I would expect the maximum area to be for a nearly circular fence)?

does the problem specify that the fence is to the right of the parabola (because it's obvious that a fence to the left would enclose a larger area)? :confused:
 
it's a rectangular (sort of) shaped fencing (except for the curved part). and the original picture had it the fencing on the right side.
 
http://img225.imageshack.us/img225/1417/86034903vj0.jpg

Forgetting (x,t)
try this one.

I have one equation in the diagram,
and you should get other two

you should get
A = ...
and constraint
maybe
x^2+x+2b = 2

Now, it's just a simple problem if you use langrange thing

del f = lamda . del g one ...
 
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the only problem is that this is a calc 1 class, and we're not permitted to use that method yet. i wish i could since a friend said it'd be easier.
 
Mylala88 said:
the only problem is that this is a calc 1 class, and we're not permitted to use that method yet. i wish i could since a friend said it'd be easier.

Forget it, use subsitution

initially, you have three equations and three unknowns.
make it to one eqn one var (to area eqn)
and differentiate the area eqn

I don't think it matters which var you choose. You would have one var in the end ..
 
Hi Mylala88! :smile:

Your integral [tex]\int_0^a (a\ -\ 4\ +\ x^2) dx[/tex] is wrong.

You've taken vertical slices of thickness dx, so their length should be … ? :smile:
 
Ah! I didn't even realize that! It's from 4 - t to 4. Thanks! I'll keep trying.
 

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