Eighty metres of fencing are available to enclose a rectangular area. What are the dimensions that enclose the max area. What is max area?

P = the perimeter = 80 m

A = the area

W = the width

L = the length

P = 2 W + 2 L = 2 ( W + L )

80 = 2 ( W + L ) Divide both sides by 2

40 = W + L Subtract L to both sides

40 - L = W + L - L

40 - L = W

W = 40 - L

A = W * L

A = ( 40 - L ) * L

A = 40 L - L ^ 2

A = - L ^ 2 + 40 L

Since the coefficient of x ^ 2 is negative, the parabola opens downward and has a maximum value at the vertex.

The vertex of the parabola is:

- b / 2 a

In this case :

a = - 1

b = 40

- b / 2 a = - 40 / [ 2 * ( - 1) ] = - 40 / - 2 = 20

That's the length of the rectangle of maximum area.

L = 20 m

We need to find the width from the equation :

W = 40 - L

W = 40 - 20

W = 20 m

So we see that the largest area possible is when the rectangle
is chosen to be the square 20 m by 20 m.

Amax = 20 * 20 = 400 m ^ 2

Ah, the exciting world of fencing! Let's see if we can help you find the optimal dimensions for your rectangular area, while having a laugh along the way.

To find the dimensions that will enclose the maximum area with your 80 meters of fencing, we need to take a stroll through geometry park. So grab your measuring tape and let's get started!

Let's assume the length of your rectangular area is L, and the width is W. Now, remember that the perimeter of a rectangle is given by the formula: P = 2L + 2W. Since you have 80 meters of fencing available, we can set up an equation: 2L + 2W = 80.

Now, let's focus on finding the maximum area for this rectangular area. The formula for the area of a rectangle is A = L * W, so we want to maximize this function.

To simplify things, let's solve the perimeter equation for L and substitute it into the area equation:

2L + 2W = 80,
L = (80 - 2W) / 2,
A = L * W = ((80 - 2W) / 2) * W.

Are you still with me? Great, because now it's time to put our funny problem-solving hats on and find the maximum area. One way to do this is by finding the vertex of a parabola, but we won't get too technical here (who wants maths to get too serious, right?).

To cut a long story short, the maximum area occurs when the width, W, is half of the available fencing, which in this case, is 40 meters. So, the optimal dimensions of your fancy rectangle are a width of 40 meters and a length of (80 - 2 * 40) / 2 = 20 meters.

Now, drumroll, please! The maximum area you can enclose with your 80 meters of fencing is 40 meters (width) multiplied by 20 meters (length), which equals 800 square meters.

Voila! You've got your optimal dimensions and max area, ready to fence in some memorable moments.

To find the dimensions that enclose the maximum area using the given 80 meters of fencing, we can use the properties of a rectangle.

Let's assume that the length of the rectangle is L meters and the width is W meters.

1. Since the perimeter is the sum of all sides, we can set up the equation:
2L + 2W = 80

Simplifying the equation:
L + W = 40 (dividing both sides by 2)

Expressing one variable in terms of the other:
L = 40 - W

2. The area of a rectangle is calculated by multiplying its length and width:
A = L * W

Substitute the value of L from the previous equation into the area equation:
A = (40 - W) * W

Simplifying the equation:
A = 40W - W^2

3. To maximize the area, we need to differentiate the equation and find its critical points. We will then determine the value of W that makes the derivative equal to zero.

dA/dW = 40 - 2W

Set the derivative equal to zero and solve for W:
40 - 2W = 0
-2W = -40
W = 20

4. Now that we have found the value of W, we can substitute it back into the equation from step 1 to find L:
L = 40 - W
= 40 - 20
= 20

So, the dimensions that enclose the maximum area are 20 meters by 20 meters.

5. The maximum area can be calculated by substituting the values of L and W into the area equation:
A = L * W
= 20 * 20
= 400 square meters

Therefore, the rectangle with dimensions 20 meters by 20 meters will enclose the maximum area of 400 square meters using the given 80 meters of fencing.

To find the dimensions that enclose the maximum area, we need to use the given information about the fencing.

Let's assume the rectangular area has a length of L meters and a width of W meters.

Now, we know that the perimeter of a rectangle is given by the formula: Perimeter = 2L + 2W.

In this case, we are given that there are 80 meters of fencing available, so we can write the equation: 2L + 2W = 80.

Now, we need to express one variable in terms of the other so that we can maximize the area. Solving the equation for one variable in terms of the other will help us do that.

To do this, let's isolate L on one side of the equation. Subtracting 2W from both sides, we get: 2L = 80 - 2W.

Dividing both sides by 2, we get: L = 40 - W.

Now we have L in terms of W. We can substitute this expression for L in the formula for the area of a rectangle: Area = Length × Width.

Substituting L = 40 - W, we get: Area = (40 - W) × W.

Expanding this equation, we have: Area = 40W - W^2.

To find the dimensions that enclose the maximum area, we need to find the value of W that maximizes the Area equation.

The maximum area occurs at the vertex of a parabola. In this case, the Area equation represents a downward-opening parabola because of the negative coefficient on the W^2 term.

The x-coordinate of the vertex of a parabola is given by the formula: x = -b / (2a), where a and b are coefficients in the quadratic equation.

In our case, the coefficient of W^2 is -1, and the coefficient of W is 40. Substituting these values into the formula, we get: W = -40 / (2 × (-1)).

Simplifying, we have: W = -40 / (-2) = 20.

So, the width that maximizes the area is 20 meters.

To find the length, we substitute this value of W back into one of the earlier equations. Using L = 40 - W, we have: L = 40 - 20 = 20.

Therefore, the dimensions that enclose the maximum area are length = 20 meters and width = 20 meters.

To find the maximum area, we substitute these values back into the area equation: Area = (40 - W) × W = (40 - 20) × 20 = 20 × 20 = 400 square meters.

So, the maximum area that can be enclosed with 80 meters of fencing is 400 square meters.

20*20 is not rectangle :D