SOLUTION: being the first grandson, your grandparents decided to give you a rectangular field for your coming wedding. If you are given 200m wires of fencing, what dimensions would you choos

Question 1185698: being the first grandson, your grandparents decided to give you a rectangular field for your coming wedding. If you are given 200m wires of fencing, what dimensions would you choose to get the maximum area?
a. List all the possible dimensions of the rectangular field.
b. Make a table of values for the possible dimensions.
c. Compute the area for each possible dimension.
d. What is the maximum area you obtained?
e. What are the dimensions of the maximum area you obtained?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this optimization problem:
**Understanding the Problem**
We have a fixed perimeter (200m of fencing) and want to maximize the area of a rectangle.
**a. Possible Dimensions:**
Let *l* be the length and *w* be the width of the rectangle. The perimeter is given by:
2*l + 2*w = 200
We can simplify this to:
l + w = 100
Since length and width must be positive, the possible dimensions are any values of *l* and *w* that satisfy this equation, where 0 < l < 100 and 0 < w < 100. It's impossible to list *all* possible dimensions because they are continuous, not discrete, but we can list some examples:
* l = 10, w = 90
* l = 20, w = 80
* l = 30, w = 70
* l = 40, w = 60
* l = 50, w = 50
* l = 60, w = 40
* l = 70, w = 30
* l = 80, w = 20
* l = 90, w = 10
**b. Table of Values:**
It's helpful to organize the dimensions and areas in a table. Here are some examples:
| Length (l) | Width (w) | Area (A) |
|---|---|---|
| 10 | 90 | 900 |
| 20 | 80 | 1600 |
| 30 | 70 | 2100 |
| 40 | 60 | 2400 |
| 50 | 50 | 2500 |
| 60 | 40 | 2400 |
| 70 | 30 | 2100 |
| 80 | 20 | 1600 |
| 90 | 10 | 900 |
**c. Computing the Area:**
The area of a rectangle is given by:
A = l * w
We've already calculated the areas in the table above.
**d. Maximum Area:**
From the table, we can see that the maximum area is 2500 m².
**e. Dimensions of Maximum Area:**
The dimensions that give the maximum area are l = 50m and w = 50m. This means the rectangle with the maximum area is a square.
**General Solution (using calculus):**
While the table gives the answer, calculus can be used for a more general approach.
1. Express *w* in terms of *l*: w = 100 - l
2. Area formula: A = l(100 - l) = 100l - l²
3. To find the maximum area, take the derivative of A with respect to *l* and set it equal to zero:
dA/dl = 100 - 2l = 0
2l = 100
l = 50
4. Since w = 100 - l, then w = 100 - 50 = 50.
This confirms that the maximum area occurs when l = 50 and w = 50.

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