You have 800 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river,find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Thank you!

How did it become 800-4w from 2400-2w^2?

Ah, the tricky art of fence optimization! Well, in this situation, we need to find the length and width that will give us the largest area with a given amount of fencing.

Now, since we're not fencing the side along the river, we only need to enclose three sides of the plot. So we have 800 feet of fencing to use for the three sides.

Let's call the width of the rectangular plot "x." So now we have two equal sides of length x and one side of length y (which is the length of the plot along the river). Since we're using a total of 800 feet of fencing, we can deduce that:

2x + y = 800

To maximize the area, we need to find the optimal values for x and y. To do that, we'll express y in terms of x:

y = 800 - 2x

Now, we can express the area of the plot, A, as a function of x:

A = x * y
A = x * (800 - 2x)

To find the maximum area, we can take the derivative of A with respect to x and set it equal to zero:

dA/dx = 800 - 4x

Setting it equal to zero gives us:

800 - 4x = 0
4x = 800
x = 200

Plugging this value back into the equation for y, we get:

y = 800 - 2 * 200
y = 400

Thus, the maximum area can be achieved when the width, x, is 200 feet and the length along the river, y, is 400 feet. Substituting these values back into the area equation:

A = 200 * 400
A = 80,000 square feet

And voila! The largest area that can be enclosed within 800 feet of fencing is 80,000 square feet. Just be sure not to clown around too much with the measurements!

To find the length and width that will maximize the area, we need to write an equation for the area of the rectangular plot.

Let's denote the length as L and the width as W. Since you have 800 feet of fencing and one side is along the river, you will only need to fence three sides of the plot: the two widths and one length.

The equation for the perimeter of the plot is:
P = 2L + W

Since you have 800 feet of fencing, we can write:
2L + W = 800

Now, we need to express the width in terms of the length. Since the side along the river does not need to be fenced, the width can be expressed as:
W = 800 - 2L

To find the area, multiply the length (L) by the width (W):
A = L * W

Substituting the value of W, we get:
A = L * (800 - 2L)

Now, we have the formula for the area (A) in terms of the length (L). To find the length that maximizes the area, we can take the derivative of A with respect to L, set it equal to zero, and solve for L.

dA/dL = 0

Applying the derivative, we get:
dA/dL = 800 - 4L

Setting the equation equal to zero, we have:
800 - 4L = 0

Solving for L, we find:
L = 200

Now, to find the corresponding width:
W = 800 - 2L
W = 800 - 2(200)
W = 400

Therefore, the length that maximizes the area is 200 feet and the width is 400 feet. To find the largest area, substitute these values into the area equation:
A = L * W
A = 200 * 400
A = 80,000 square feet

Thus, the largest area that can be enclosed is 80,000 square feet.

I know how to do this with calculus, but algebraically it starts the same way.

L + 2W = 800 L = 800 - 2w
A = LW

A= (800-2w)w

A = 800w - 2w^2

Take the derivative: 800 -4w
800 -4w = 0 to find max.
800 = 4w
200 =w
l = 400