1. A gardener has 140 feet of fencing to fence in a rectangular vegetable garden. Find the dimensions of the largest area he can fence. Find the possible rectangular area he can enclose.

2. Suppose a farmer has a large piece of land and he wants to make a rectangular fence for his animals, but he has no money to buy more wood for the fence. The total length of the fence is fixed to 250 meters. What should be the width and length of the rectangle such that the area is maximized?

Hints:

It can be shown by calculus that the maximum area of a polygon enclosed by a fixed length of perimeter is a regular polygon, i.e. a polygon with all sides and interior angles equal.

It follows from the above that the maximum area enclosed by a fixed perimeter is a circle, equivalent to a polygon with "infinite" number of sides.

1. To find the dimensions of the largest area, we can use the formula for the area of a rectangle:

A = length × width

Given that the gardener has 140 feet of fencing, we can set up the equation:
2 × length + 2 × width = 140

Simplifying the equation, we get:
length + width = 70

Next, we can rearrange the equation to solve for either length or width. Let's solve for length:
length = 70 - width

Substitute this value of length into the area formula:
A = (70 - width) × width

To find the maximum area, we need to find the value of width that maximizes the area. We can do this by finding the vertex of the quadratic equation:
A = -width^2 + 70 × width

The vertex of a quadratic equation in the form A = ax^2 + bx + c is given by:
x = -b / (2a)

In this case, a = -1, b = 70, and c = 0. Substituting these values, we get:
width = -70 / (2 × (-1))
width = 35

Substituting this value of width back into the equation for length, we get:
length = 70 - 35
length = 35

Therefore, the dimensions of the largest area the gardener can fence are 35 feet by 35 feet, and the possible rectangular area he can enclose is 35 feet × 35 feet.

2. To find the dimensions of the rectangle that maximizes the area when the total length of the fence is fixed to 250 meters, we can follow a similar approach.

Let the width of the rectangle be w and the length be l.

We are given that the total length of the fence is 250 meters, so we can set up the equation:
2 × l + 2 × w = 250

Simplifying the equation, we get:
l + w = 125

Next, we can rearrange the equation to solve for either l or w. Let's solve for l:
l = 125 - w

Substitute this value of l into the area formula:
A = (125 - w) × w

To find the maximum area, we need to find the value of w that maximizes the area. We can do this by finding the vertex of the quadratic equation:
A = -w^2 + 125 × w

The vertex of a quadratic equation in the form A = ax^2 + bx + c is given by:
x = -b / (2a)

In this case, a = -1, b = 125, and c = 0. Substituting these values, we get:
w = -125 / (2 × (-1))
w = 62.5

Substituting this value of w back into the equation for l, we get:
l = 125 - 62.5
l = 62.5

Therefore, the width should be 62.5 meters and the length should be 62.5 meters to maximize the area.

To find the dimensions of the largest area that can be fenced using a given amount of fencing, we can use the concept of optimization.

1. The gardener has 140 feet of fencing, which means the total length of the fence will be 140 feet.
Let's assume the width of the rectangular vegetable garden is w feet, and the length is l feet.

To maximize the area, we need to express the area of the rectangle in terms of a single variable. We can use the fact that the perimeter of a rectangle is equal to twice the sum of its width and length.

Perimeter = 2w + 2l = 140 feet
Simplifying the equation, we have w + l = 70 feet

Now, since we want to maximize the area, we need to express the area as a function of a single variable.
The area of a rectangle is given by A = w * l.

We can substitute w = 70 - l into the area equation to express the area in terms of a single variable:
A = (70 - l) * l = 70l - l^2

To find the dimensions of the largest area, we need to maximize the function A = 70l - l^2.

To do this, we can take the derivative of A with respect to l and set it equal to zero, and then solve for l to find the critical points.

dA/dl = 70 - 2l = 0
Solving this equation, we find l = 35 feet.

To determine if this value of l gives us a maximum or minimum, we can take the second derivative of A with respect to l and evaluate it at the critical point.

d^2A/dl^2 = -2
Since the second derivative is negative, we can conclude that l = 35 feet gives us the maximum area.

Now, we can substitute this value of l back into our equation w + l = 70 feet to find the width:
w + 35 = 70
w = 70 - 35
w = 35 feet

So, the dimensions of the largest area that can be fenced in with 140 feet of fencing are 35 feet by 35 feet. The maximum area that can be enclosed is 35 * 35 = 1225 square feet.

2. In this case, the total length of the fence is fixed at 250 meters. Let's assume the width of the rectangular fence is w meters, and the length is l meters.

The perimeter of a rectangle is equal to twice the sum of its width and length.
Perimeter = 2w + 2l = 250 meters

To express the area of the rectangle in terms of a single variable, we can use the fact that the perimeter is fixed.
w + l = 125 meters

The area of a rectangle is given by A = w * l.

We can substitute w = 125 - l into the area equation to express the area in terms of a single variable:
A = (125 - l) * l = 125l - l^2

To find the dimensions that maximize the area, we need to maximize the function A = 125l - l^2.

Taking the derivative of A with respect to l and setting it equal to zero, we can solve for l to find the critical points.

dA/dl = 125 - 2l = 0
Solving this equation, we find l = 62.5 meters.

To determine if this value of l gives us a maximum or minimum, we can take the second derivative of A with respect to l and evaluate it at the critical point.

d^2A/dl^2 = -2
Since the second derivative is negative, we can conclude that l = 62.5 meters gives us the maximum area.

Substituting this value of l back into w + l = 125 meters, we can find the width:
w + 62.5 = 125
w = 125 - 62.5
w = 62.5 meters

So, the dimensions of the rectangle that maximize the area with a fixed perimeter of 250 meters are 62.5 meters by 62.5 meters. The maximum area that can be enclosed is 62.5 * 62.5 = 3906.25 square meters.