find largest area possible given 200 feet of fencing, n ft

C

cabriosnob

New member
Joined
Oct 8, 2008
Messages
7
Joshua has decided to convert part of his back yard into a garden. He knows from his neighbor’s experience that he needs to fence in the garden to keep the animals out. At the hardware store, he buys 200 feet of chicken wire. After he gets home and unrolls the wire, he finds to his surprise that there are many different size gardens that he can make with 200 feet of fencing. He decided that he wants to get the largest garden from 200 feet of fencing. What are the dimensions of his garden? What if he bought n feet of wire?

I am horrible at these kinds of problems. I have graph paper, I have drawn a 50x50 square, a 30x20x30x20x30x20x30x20 octagon, and a 50x50 rhombus. I don't really know where to start, how to find a solution to the specific question regarding 200 feet, or how to find and algebraic expression for the n feet of wire. I am completely lost, can't get started, can't figure anything out.

Can you please get me started and help me find the general equation?
 
cabriosnob said:
Can you please get me started and help me find the general equation?
Click to expand...
Under the stated conditions (any shape being allowed), there is, to my knowledge, no "general equation". You may need to use higher-level techniques such as calculus to prove the shape that has the largest area for a given perimeter. :shock:

Eliz.
 
Re: area and perimeter

Circle has the smallest perimeter for a given area.
 
Re: area and perimeter

This problem doesn't require calculus or anything very complicated, that's what's driving me nuts. It's for a Geometry for Teachers class. It's a problem-solving report. She always gives us a sepcific probem then asks for a general solution. They are always easily solved and always have an algebraic general solution. My problem is I always make them harder than they have to be and can't get started.
 
cabriosnob said:
This problem doesn't require calculus or anything very complicated...
Click to expand...
Actually, as the discussion in the link provided earlier explained, yes, you do indeed need calculus, etc, to answer the question as posted. :oops:

For this to have a "simpler" proof, there must have been some other conditions, not included in your post, which put some sort of restrictions on the figures to be considered. Please review the assignment, or consult with your instructor. Thank you! :D

Eliz.
 
Re: area and perimeter

Let's suppose you want a rectangular garden and you don't want to use calculus. If this is the case, you have to consider two formulas... Area = length times width and Perimeter = 2 times the length + 2 times the width.
You know the perimeter to be 200.
Let's be sure to identify stuff. Take the time to write it out.

Let x represent the length.
Let y represent the width.

Now, if you solve for one of these (x or y) in terms of the perimeter and then substitute that value into the area formula you can graph the resulting equation. (The vertical axis is the area. The horizontal axis is either the length or the width.) You will probably get a cup-down parabola. Find the coordinates of the highest point on the parabola and you have the length or width, depending on whether you found x or y in your first maneuver.
 
Loren said:
Let's suppose you want a rectangular garden and you don't want to use calculus.
Click to expand...
Well, sure: With the restriction of "a rectangular garden", you don't need calculus. :wink:

Eliz.
 
Re: area and perimeter

Eliz.,
Your links with calculus would be helpful if this were not a clas for teaching to elementary school students. This problem does not require such difficult concepts. There were some non-calculus responses that were extremely helpful. And yes, I did include the entire problem. This assignment was given in tandem with the lesson on introducing area and perimeter, I have to solve it and then provide elementary level extensions and modifications. The idea is that this can be solved by 3rd and 4th graders. Calculus does not apply. Geoboards and manipulatives will help solve this problem. It's an elementary school problem, I just can't get started. I have a rhombus so far as enclosing the most area, but I don't think that's right.

The teacher is trying to get us to think in different ways and to find a pattern and make a general equation in n. We do these all the time and they are very easy. I'm just stuck on this one.
 
Re: area and perimeter

Loren and Subhotosh,
Thanks!!! That's exactly what I needed!!! I can't believe I made such a mountain out of something so simple! I really appreciate your time and help!!!

Eliz.,
Thanks for your time and help too, I understand what you were getting at...
 
Re: area and perimeter

cabriosnob. Try the following investigations.

Consider a rectangle having a perimeter of say 16.
Now consider the following lengths and widths that produce the area.

length X width=Area
1 X 7 = 7
2 X 6 = 12
3 X 5 = 15
4 X 4 = 16
5 X 3 = 15
6 X 2 = 12
7 X 1 = 7

It seems quite obvious you get a maximum area when the length and width of the rectangle are equal, producing a square. Regarding your rhombus. Start with a square. Squeeze two opposite corners toward each other to form a rhombus without altering the lengths of the sides. Do you see what happens to the area when you squeeze it so that one pair of opposite corners of the square are squeezed together?

You can do similar things with a triangle, pentagon, etc. I think you will discover that you get the maximum area when you have a regular polygon. Now, consider a polygon with an infinite number of sides. I guess you get pretty close to a circle.

Good luck in your experimentations.
 
cabriosnob said:
This assignment was given in tandem with the lesson on introducing area and perimeter...
Click to expand...
Ah, so there was additional information: The exercise was to be completed within the context of simple polygons only, probably using nothing more complex than rectangles.

Thank you! :D

Eliz.
 
Re: area and perimeter

Joshua has decided to convert part of his back yard into a garden. He knows from his neighbor’s experience that he needs to fence in the garden to keep the animals out. At the hardware store, he buys 200 feet of chicken wire. After he gets home and unrolls the wire, he finds to his surprise that there are many different size gardens that he can make with 200 feet of fencing. He decided that he wants to get the largest garden from 200 feet of fencing. What are the dimensions of his garden? What if he bought n feet of wire?

I am horrible at these kinds of problems. I have graph paper, I have drawn a 50x50 square, a 30x20x30x20x30x20x30x20 octagon, and a 50x50 rhombus. I don't really know where to start, how to find a solution to the specific question regarding 200 feet, or how to find and algebraic expression for the n feet of wire. I am completely lost, can't get started, can't figure anything out.

Can you please get me started and help me find the general equation?

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding
"a" to one side and subtracting "a" from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
Therefore, the dimensions of the garden with maximum area for the given perimeter, p, of fencing is p/4 by p/4.
Given a length of fencing of 200 feet, the garden dimensions become by 50 x 50 ft. for an area of 2500 sq.ft.

Draw yourself polygons with 4, 6, 8, 10, etc. sides with perimeters of 200 and compute their areas. You will soon see that the area approaches that of the circle of circumference 200.

Consider a polygon with "n" sides of length "s". Connecting the vertices with the center produces "n" triangles with altitudes "a".
The area of each triangle is A = as/2. The area of all "n" triangles is A = ans.

The maximum area garden for a given perimeter is a circle.

As "n" increases, the area increases,

As "n" increases, "a" increases and "s" decreases, ultimately resulting in the circle of radius r = a and "ns" becoming C = 2(Pi)r.

Therefore, the area of the figure becomes A = r(2)Pi(r)/2 = Pir^2, the figure with maximum area for a given perimeter.
Then, A = Pir^2 where r = P x Pi making A = 3183 sq.ft.
 
You must log in or register to reply here.
Top