a. A rectangular pen is built with one side against a barn. 1200


m of fencing are used for the other three sides of the pen. What dimensions maximize the area of the​ pen?
b. A rancher plans to make four identical and adjacent rectangular pens against a​ barn, each with an area of 25

msquared
.
What are the dimensions of each pen that minimize the amount of fence that must be​ used?

(a) as with all these, divide the fence so that the lengths (1) equal the widths(2). Thus the pen is 600x300

(b) same problem, but now there are 4 widths, so the same principal applies. To minimize the fence, each pen has dimensions x and 25/x, where 4x = 5(25/x). That is, x^2 = 125/4, or x = 5√5/2

each pen is 5√5/2 by 2√5, with area of 25.

More generally, the fencing needed is
f = 4x + 5(25/x)
df/dx = 4 - 125/x^2
for minimum fencing, df/dx = 0, so x^2 = 125/4 as above.

To find the dimensions that maximize the area of the rectangular pen in question (a), we can use the concept of optimization.

First, let's define the variables we need:
- Let the width of the pen be x meters.
- The length of the pen, parallel to the barn, will be 1200m - 2x (since one side is against the barn).

Now, we can find the area (A) of the pen in terms of x:
A = x * (1200 - 2x)
A = 1200x - 2x^2

Next, we need to maximize the area, which means finding the maximum value of A. To do this, we can take the derivative of A with respect to x and set it equal to zero:

dA/dx = 1200 - 4x
1200 - 4x = 0
4x = 1200
x = 300

By plugging this value of x into the equation for the length, we can find the dimensions that maximize the area:
Length = 1200 - 2x = 1200 - 2 * 300 = 1200 - 600 = 600 meters
Width = x = 300 meters

Therefore, the dimensions that maximize the area of the rectangular pen are 600 meters for the length and 300 meters for the width.

Now, let's move on to question (b) where we need to find the dimensions that minimize the amount of fence used for four rectangular pens.

Let's define the variables again:
- Let the width of each pen be x meters.
- The length of each pen, parallel to the barn, will be (25/x) meters since the area of each pen is 25 m^2 (length * width = 25).

To find the amount of fence used for one pen, we add up the lengths of all the sides:
Fence used for one pen = 2x + (25/x) + (25/x) + x = 2x + 2(25/x) + x = 3x + 50/x

Since there are four pens, the total amount of fence used will be four times the above expression:
Total fence used = 4 * (3x + 50/x) = 12x + 200/x

To minimize the fence used, we once again take the derivative of the above expression with respect to x and set it equal to zero:

d(12x + 200/x)/dx = 12 - 200/x^2
12 - 200/x^2 = 0
200/x^2 = 12
200 = 12x^2
x^2 = 200/12
x^2 = 50/3
x = sqrt(50/3)
x ≈ 3.54

By plugging this value of x into the expression for the length, we can find the dimensions that minimize the fence used:
Length = 25/x ≈ 25/(sqrt(50/3)) ≈ 8.84 meters
Width = x ≈ 3.54 meters

Therefore, the dimensions of each pen that minimize the amount of fence used are approximately 8.84 meters for the length and 3.54 meters for the width.

a. To maximize the area of the rectangular pen, we need to find the dimensions that will give us the largest possible area.

Let's assume the width of the pen is 'w' and the length is 'l'.

According to the given information, the pen has one side against a barn, which means the length of the pen is connected to the barn. Hence, the width will be the only variable side that affects the area.

The perimeter of the pen is given as 1200m, which means the sum of all three sides (excluding the side connected to the barn) is 1200m. We can represent this as:

2w + l = 1200

Now, the area of the pen can be calculated as:

Area = length * width
Area = l * w

To get the dimensions that maximize the area, we need to express the area in terms of a single variable. This can be done by solving the perimeter equation for 'l' and then substituting it into the area equation.

From the perimeter equation:
l = 1200 - 2w

Substituting l in the area equation:
Area = (1200 - 2w) * w
Area = 1200w - 2w^2

To find the maximum area, we need to find the value of 'w' that maximizes the quadratic equation. This can be achieved by finding the vertex of the quadratic equation.

The vertex can be found using the formula: w = -b / 2a, where 'a' is the coefficient of the quadratic term (-2w^2) and 'b' is the coefficient of the linear term (1200w).

In this case, a = -2 and b = 1200.

w = -1200 / 2(-2)
w = -1200 / -4
w = 300

So, the width that maximizes the area is 300m. To find the corresponding length, we can substitute this value of 'w' into the perimeter equation:

2w + l = 1200
2(300) + l = 1200
l + 600 = 1200
l = 1200 - 600
l = 600

Therefore, the dimensions that maximize the area of the rectangular pen are a width of 300m and a length of 600m.

b. To minimize the amount of fence that must be used for four identical and adjacent rectangular pens, each with an area of 25m^2, we need to find the dimensions that minimize the perimeter of each pen.

Let's assume the width of each pen is 'w' and the length is 'l'.

The area of each pen is given as 25m^2, so we can write:

Area = length * width
25 = l * w

To minimize the perimeter, we can express it in terms of a single variable. The perimeter of each pen is given by:

Perimeter = 2(length + width)
Perimeter = 2(l + w)

We want to minimize the perimeter while keeping the area at 25m^2.

Substituting the area equation into the perimeter equation, we get:

Perimeter = 2(l + 25/l)

To minimize this expression, we take the derivative with respect to 'l' and set it equal to zero. Then, we solve for 'l'.

d(Perimeter)/dl = 0
2(1 - 25/l^2) = 0
1 - 25/l^2 = 0
25/l^2 = 1
l^2 = 25
l = 5

Now, we can substitute the value of 'l' into the area equation to find the corresponding width:

Area = l * w
25 = 5 * w
w = 5

Therefore, each pen will have dimensions of 5m by 5m to minimize the amount of fence that must be used.