A rancher wants to build a rectangular fence next to a river, using 100 yd of fencing. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the rancher should not fence the side next to the river.)

If the length (parallel to the river) is y,

2x+y = 100
a = xy = x(100-2x) = 100x-2x^2
max a at x = 100/4 = 25

So, the yard is 25x50

As usual, max area when the fence is divided equally among lengths and widths.

To find the dimensions of the rectangle that will maximize the area, we need to set up an equation and then differentiate it to find its maximum value.

Let's assume the length of the rectangle (along the river) is L, and the width (perpendicular to the river) is W.

From the question, we know the rancher wants to use 100 yd of fencing. There are four sides to the rectangle, and the side next to the river does not require fencing. Therefore, the amount of fencing used is equal to the perimeter of the rectangle minus the length next to the river, which is L:

2L + W = 100 (equation 1)

We need to express the area of the rectangle in terms of L and W. The area (A) of a rectangle is calculated as the product of its length and width:

A = L * W (equation 2)

Now, we need to express W in terms of L so that we can plug it into equation 2 and have A expressed solely in terms of L:

From equation 1, we can solve for W:

W = 100 - 2L

Now we can substitute this expression for W back into equation 2:

A = L * (100 - 2L) = 100L - 2L^2 (equation 3)

To find the maximum area, we need to find the value of L that maximizes equation 3. To do that, we differentiate equation 3 with respect to L, and then set the resulting derivative equal to zero:

dA/dL = 100 - 4L

Setting dA/dL equal to zero and solving for L:

100 - 4L = 0
4L = 100
L = 25

So, L = 25 is the value that maximizes the area. Now, we can substitute this value back into equation 3 to find the maximum area:

A = 100L - 2L^2
A = 100(25) - 2(25)^2
A = 2500 - 2(625)
A = 2500 - 1250
A = 1250

Therefore, the dimensions that will maximize the area of the rectangle are L = 25 yd and W = 100 - 2L = 50 yd, and the maximum area is 1250 square yards.

To find the dimensions that will maximize the area, we can start by setting up the problem.

Let's assume the length of the fence is represented by 'L' and the width is represented by 'W'. Since the rancher is only fencing three sides of the rectangle, the length (L) will be parallel to the river, while the width (W) will be perpendicular to the river.

We know that the total amount of fencing available is 100 yards, and we need to find the dimensions that maximize the area.

Step 1: Set up the equations
Since the length (L) needs to be parallel to the river, and there are two widths (W), the total length of fencing used can be represented as:
2W + L = 100

We also know that the area (A) of the rectangle is given by:
A = L * W

Step 2: Solve for L in terms of W
From the equation 2W + L = 100, we can solve for L:
L = 100 - 2W

Step 3: Substitute L in the area equation
Substituting the value of L in terms of W into the area equation, we get:
A = (100 - 2W) * W

Step 4: Find the critical point
To find the maximum area, we need to find the critical point. We do this by finding where the derivative of the area equation is equal to zero.

Let's differentiate A with respect to W:
dA/dW = 100 - 4W

Set the derivative equal to zero and solve for W:
100 - 4W = 0
4W = 100
W = 25

Step 5: Find the maximum area
To find the maximum area, substitute W = 25 back into the area equation:
A = (100 - 2W) * W
A = (100 - 2 * 25) * 25
A = 50 * 25
A = 1250 square yards

So, the dimensions that will maximize the area are: Length (L) = 100 - 2W = 100 - 2(25) = 50 yards, Width (W) = 25 yards. The maximum area is 1250 square yards.