Word Problems:

1. A farmer has 2400 feet of fencing and wants to fence off a rectangular field that borders a straight river. If he needs no fence along the river, what are the dimensions of the field that he can fence with the largest area? Find the area of the field.

2. If 1800 feet of fencing is available to build five adjacent pens, express the total area of the pens in terms of x. What value of x will maximize the total area?

#1. x(2400-2x)

#2. x(1800-6x)/2

Both of these are just parabolas, where the maximum area occurs at the vertex of the parabola.

Recall that for y = ax^2+bx+c, the vertex is at x = -b/2a

1. To find the dimensions of the field that will result in the largest area, we will use the fact that the perimeter of the rectangular field is equal to 2400 feet. Let's call the length of the field L and the width of the field W.

The perimeter of a rectangle is given by the formula: P = 2L + 2W.

In this problem, we know the perimeter is 2400 feet, so we can write the equation as:
2400 = 2L + 2W.

Now, we need to express one of the variables in terms of the other so we can represent the area in terms of a single variable.

Solving the equation above for L, we get:
2L = 2400 - 2W,
L = 1200 - W.

The area of a rectangle is given by the formula: A = L * W.

Substituting L = 1200 - W into the area formula, we have:
A = (1200 - W) * W.

To find the dimensions that maximize the area, we can find the value of W that maximizes A. Since the field cannot have negative dimensions, we know that 0 ≤ W ≤ 1200.

To maximize A, we take the derivative of A with respect to W and set it equal to 0. Let's differentiate A with respect to W:

dA/dW = 1200 - 2W.

Setting dA/dW = 0, we have:
1200 - 2W = 0,
2W = 1200,
W = 600.

Plugging this value of W back into the equation for L, we have:
L = 1200 - 600,
L = 600.

Therefore, the dimensions of the field that will result in the largest area are: length = 600 feet, width = 600 feet.

To find the maximum area, substitute these dimensions into the area formula:
A = L * W = 600 * 600 = 360,000 square feet.

So, the farmer can fence off a rectangular field with dimensions 600 feet by 600 feet, resulting in the largest area of 360,000 square feet.

2. Let's call the length of each pen L and the width of each pen W. We know that the total length of the fencing available is 1800 feet and that there are five adjacent pens, so we can write the equation:

1800 = 5L + 10W.

To express the total area of the pens in terms of x, we need to find the relationship between L, W, and x.

Since there are five adjacent pens, the total length of the pens is equal to 5L. Now, the value of W will depend on x.

The length of each pen is divided into two parts: x and L - x (since we want x to be the length of the pen, and the remaining length will be L - x).

Since there are five adjacent pens, the total length is 5L, and each pen has length x + L - x = L. So, we can write the equation for the total length as:

5L = 1800 - 10W.

To express W in terms of x, we need to solve for W:

10W = 1800 - 5L,
W = (1800 - 5L)/10,
W = (1800 - 5(x + L - x))/10,
W = (1800 - 5L)/10.

The area of each pen is given by the formula: A = L * W.

Substituting W = (1800 - 5L)/10 into the formula for the area, we have:

A = L * [(1800 - 5L)/10].

To find the value of x that maximizes the total area A, we differentiate A with respect to L and set it equal to 0. Let's differentiate A with respect to L:

dA/dL = (1800 - 5L)/10 - (5L)/10,
dA/dL = (1800 - 10L - 5L)/10,
dA/dL = (1800 - 15L)/10.

Setting dA/dL = 0, we have:
1800 - 15L = 0,
15L = 1800,
L = 120.

Now, we can find W and A by substituting L = 120 into the respective equations:

W = (1800 - 5L)/10,
W = (1800 - 5(120))/10,
W = 180 feet.

A = L * W,
A = 120 * 180,
A = 21,600 square feet.

So, the value of x that maximizes the total area of the pens is 120 feet, and the maximum total area is 21,600 square feet.

1. To find the dimensions of the field that can be fenced with the largest area, we need to use the concept of derivative and optimization.

Let's breakdown the problem. We have a rectangular field that borders a straight river, so we only need three sides of fencing. Let's assume the length of the field is L and the width is W.

We are given that the farmer has 2400 feet of fencing, and we know that the field requires 3 sides of fencing, which means 2L + W = 2400.

To find the dimensions that maximize the area, we need to express the area in terms of one variable. The area of a rectangle is given by A = L * W.

Now, we need to express one variable in terms of the other. From the equation 2L + W = 2400, we can solve for W in terms of L: W = 2400 - 2L.

Substituting this value of W into the area equation, we have A = L * (2400 - 2L).

To find the maximum area, we need to find the critical points of the area function. Taking the derivative of the area function with respect to L, we get dA/dL = 2400 - 4L.

Setting dA/dL = 0 and solving for L, we have 2400 - 4L = 0. Solving this equation, we find L = 600.

Now, we can substitute this value of L back into the equation for W: W = 2400 - 2L = 2400 - 2(600) = 1200.

Therefore, the dimensions of the field that can be fenced with the largest area are L = 600 feet and W = 1200 feet.

To find the maximum area, substitute these values into the area equation: A = L * W = 600 * 1200 = 720,000 square feet.

2. In this problem, we have 1800 feet of fencing available to build five adjacent pens. Let's assume each pen has a width of x. Since the pens are adjacent, the total length of the pens will be 5x.

We know that the total fencing required will be the sum of all the sides of the pens. For each pen, we have two sides of width x and one side of length 5x. Therefore, the total fencing required is 2x + 2x + 5x = 9x.

We are given that the total fencing available is 1800 feet, so we can set up the equation 9x = 1800.

Solving this equation, we find x = 200.

Therefore, each pen will have a width of 200 feet. To find the total area of the pens, we need to multiply the width of each pen by the length of each pen, which is 5x.

Total area = 5x * x = 5 * 200 * 200 = 200,000 square feet.

To maximize the total area, we need to maximize the value of x. In this case, x = 200 will maximize the total area.