If you have 280 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Area A=x.y

The perimeter of a rectangle is 2x+y= 280 solving for y give you,m y=280-2x, y is greater to zero.
A=x*y substituting y in the area equation give you
A=x(280-2x.)=280x-2x^2
The area is maximized when A'=0
so first find the derivative of A
A'=280-4x. A=0 when 280-4x=0 solve for x. x=280/4=70 so x=70
you subtitue x into the y equation which is y=280-2x so y = 280-2(70)=280-140=140
so y=140
Finally the dimensions are x=70, and y=140 so the area is 70*140=9800m^2

Let the perimeter be 2w + 280-2w = 280

A = w * (280-2w)
A = 280w - 2w^2
dA/dw = 280 - 4w

A is max when dA/dw = 0

Thus, w=70

The dimensions are 70x140 and the area is 9800

Ah, fencing conundrums! Well, I must say, the largest area you can enclose with 280 meters of fencing against a wall is quite rectangularly amusing. Now, let's put on our math clown noses and solve this puzzle!

To maximize the area, we need to make a rectangular enclosure. Since one side is against a wall, we only need to surround three sides. Let's call the length of the wall side "x" meters and the other two sides "y" meters each.

Now, since we have 280 meters of fencing in total, it means that the perimeter of our enclosure is: 2y + x = 280.

To make things more interesting, let's solve this equation for "x" and express it in terms of "y". That yields x = 280 - 2y.

Now, the area of our rectangular enclosure is given by A = xy. Substituting the value of "x" from earlier, we have A = (280 - 2y)y.

Now, my comical friend, let's find the maximum area by finding the value of "y" that makes A as large as possible. We can do this by taking the derivative of A with respect to "y" and setting it to zero. But don't fret, I'll spare you from too many number jokes!

Taking the derivative of A = (280 - 2y)y and setting it to zero, we get 280 - 4y = 0. Solving for "y", we find y = 70.

So, the maximum area you can enclose is when both sides, not against the wall, measure 70 meters. That means the length along the wall will be 280 - 2(70) = 140 meters.

Voila! The largest area you can enclose is a rectangular field with dimensions 70 meters by 140 meters. Happy fencing!

To find the largest area you can enclose with 280 meters of fencing, you need to determine the dimensions of the rectangular area that would maximize it.

Let's assume the length of the rectangular area is L, and the width is W. Since the rectangular area is up against a long, straight wall, it only requires three sides to be fenced (i.e., two equal sides and one side against the wall).

Since there are two equal sides, the fencing used for these sides would be 2L. The remaining fencing would be used for the side against the wall and is denoted by W.

Given that the total fencing available is 280 meters, we can write the equation:

2L + W = 280 meters ---(1)

Now, we need to express the area, A, in terms of L and W. The area of a rectangle is given by A = L * W.

Therefore, the area A can be expressed as:

A = L * W ---(2)

To maximize the area, we need to find the value of L and W that satisfy equation (1) and maximize the expression (2).

Solving equation (1) for W, we get:

W = 280 - 2L

Substituting this value of W into equation (2), we get:

A = L * (280 - 2L)

Expanding and rearranging the equation:

A = 280L - 2L^2

Now, we can find the maximum area by finding the value of L that maximizes this quadratic equation.

To do this, we can take the derivative of A with respect to L and set it equal to zero:

dA/dL = 280 - 4L = 0

Solving this equation for L, we get:

280 - 4L = 0
4L = 280
L = 280/4
L = 70

Substituting this value of L back into equation (1) to find W:

2(70) + W = 280
140 + W = 280
W = 280 - 140
W = 140

So, the dimensions of the rectangular area that would maximize the area are L = 70 meters and W = 140 meters.

Finally, substituting these values into equation (2) to find the maximum area:

A = L * W
A = 70 * 140
A = 9800 square meters

Therefore, the largest area that can be enclosed with 280 meters of fencing is 9800 square meters.

To find the largest area that can be enclosed with a given amount of fencing, we can start by understanding the problem and applying some mathematical principles.

In this case, we have a long, straight wall that serves as one side of the enclosure. Let's denote the length of the wall as L (in meters). The remaining three sides of the enclosure will be perpendicular to the wall and have the same length, which we'll denote as W (also in meters).

Now, the perimeter of the enclosure is given as 280 meters, which means we can express this as an equation:
2L + 3W = 280

Since the wall accounts for one length (L), it will only appear once in the equation. The other two lengths (W) will appear twice because there are two such sides in the enclosure.

To find the largest area, we need to maximize the enclosed area defined as L × W. We can rewrite this equation as:
Area = L × W

To simplify, we need to rewrite one of the variables in terms of the other. Solving the perimeter equation for L in terms of W gives us:
L = (280 - 3W) / 2

Now, we can substitute this expression for L into our area equation:
Area = [(280 - 3W) / 2] × W

Expanding and rearranging this equation, we get:
Area = (280W - 3W^2) / 2

To find the maximum area, we take the derivative of the area equation with respect to W and set it equal to zero. Differentiating and simplifying, we have:
d(Area)/dW = 280/2 - 3W = 0

Simplifying further, we find:
140 - 3W = 0

Solving for W, we find W = 140/3 ≈ 46.67 meters.

To find the corresponding length (L), we substitute the obtained value of W back into the perimeter equation:
2L + 3(46.67) = 280
2L + 140 = 280
2L = 280 - 140
2L = 140
L = 70 meters

Therefore, to enclose the largest area with 280 meters of fencing, we should choose a rectangular enclosure with dimensions approximately 46.67 meters by 70 meters, resulting in an area of approximately 3267.67 square meters.