Max Area Problem: Find Rectangle w/ Largest Area (400 yrs Fencing)

In summary, David has 400 yards of fencing and wants to enclose a rectangular area. The area of the rectangle, A, can be expressed as a function of the width, w, as A = w(400-2w). The largest area is achieved when w = 200 yards, and the maximum area is 40,000 square yards. The teacher does not want the use of derivatives, but the perimeter formula can be used to find the value of the length, L. Using the fact that perimeter = 400 yards, we can set up the equation 2w + 2L = 400 and solve for L. However, using derivatives can also be used to check the answer.
  • #1
Cstreet09
2
0
1. David has 400 yrs of fencing and wishes to enclose a rectangular area. a) express the area A of the rectangle as a function of the width w of the rectangle. b) For what value of w is the area largest. c) What is the maximum area?



2. A=xy... My teacher does not want us to use derivatives.



3. I can't even being to attempt it.
 
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  • #2
You got A=Lw and you want the area in terms of w, so the next step is to figure out what L is (in terms of w) and sub it into A=Lw.

Hint: use the perimeter formula
 
  • #3
So does that mean L=400-w?
 
  • #4
Cstreet09 said:
So does that mean L=400-w?

No. Draw a picture and label all 4 sides L or w. What is it that equals 400?
 
  • #5
It may help to think of it this way:

What is greater?

(x-n)(x+n)
or
x2

To help, what is another way of writing the first one?


What's to say you can't use optimisation (derivatives) to check your answer?
 
  • #6
There would be no reason to use derivatives. The function to be optimized is a quadratic and completing the square works nicely.
 
  • #7
Cstreet09 said:
David has 400 yrs of fencing
That's going to take him a very long time!:wink:
 
  • #8
:wink: He probably means light-years! :biggrin:
 
  • #10
Nah, I would have thought dark years
 
  • #11
ah … dark time

that mysterious phenomenon which is generally interspersed among ordinary time, but occasionally is more concentrated, causing effective time to pass more slowly and more heavily. :wink:
 

FAQ: Max Area Problem: Find Rectangle w/ Largest Area (400 yrs Fencing)

1. What is the "Max Area Problem"?

The Max Area Problem is a mathematical problem that involves finding the rectangle with the largest area given a fixed perimeter. In this case, the perimeter is 400 yards of fencing.

2. Why is the Max Area Problem important?

The Max Area Problem has real-world applications in fields such as engineering, architecture, and economics. It helps determine the most efficient use of materials and resources to achieve the maximum area within a given perimeter.

3. How do you solve the Max Area Problem?

To solve the Max Area Problem, you need to use calculus to find the maximum value of the area function. This involves finding the derivative of the area function and setting it equal to zero to find the critical points. Then, you can use the second derivative test to determine if the critical points are maximum points.

4. What are the steps to solve the Max Area Problem?

The steps to solve the Max Area Problem are:
1. Identify the given perimeter.
2. Write the area function in terms of one variable.
3. Use calculus to find the derivative of the area function.
4. Set the derivative equal to zero and solve for the critical points.
5. Use the second derivative test to determine if the critical points are maximum points.
6. Plug the critical points into the area function to find the maximum area.

5. What is the solution to the Max Area Problem with a perimeter of 400 yards?

The solution to the Max Area Problem with a perimeter of 400 yards is a square with each side measuring 100 yards. This results in an area of 10,000 square yards, which is the maximum area achievable with a perimeter of 400 yards.

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