A farmer wants to set up a pigpen using 40 feet of fence to enclose a rectangular area of 51 square feet. Find the dimensions of the pigpen.

L w = 51

2L + 2 w = 40 so L+w=20

51/w + w = 20

w^2 - 20 w + 51 = 0

w = [ 20 +/- sqrt (400 - 204)]/2

= [ 20 +/- 14]/2
so
3 by 17

To find the dimensions of the pigpen, we need to use the given information.

Let's assume the length of the rectangular area is L feet, and the width is W feet.

We know that the perimeter of a rectangle is calculated by adding all four sides together, which is given by the equation:

Perimeter = 2L + 2W

In this case, the perimeter is given as 40 feet. So, we can write:

2L + 2W = 40 -- (Equation 1)

Also, we're given that the area of the rectangular area is 51 square feet, which is calculated by multiplying the length and width:

Area = L * W

So, we can write this equation as:

L * W = 51 -- (Equation 2)

Now, we have a system of two equations with two variables (L and W). We can use these equations to solve for L and W.

Let's solve the system using substitution. We'll solve Equation 2 for L and substitute it into Equation 1.

From Equation 2:

L = 51 / W

Substituting this into Equation 1:

2(51 / W) + 2W = 40

Now, we can simplify this equation further.

Multiply both sides by W to eliminate the fraction:

2(51) + 2W^2 = 40W

102 + 2W^2 = 40W

Rearrange the equation to get a quadratic equation:

2W^2 - 40W + 102 = 0

Now, we can solve this equation for W using the quadratic formula:

W = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 2, b = -40, and c = 102.

Substituting the values:

W = (-(-40) ± sqrt((-40)^2 - 4(2)(102))) / 2(2)

Simplifying:

W = (40 ± sqrt(1600 - 816)) / 4

W = (40 ± sqrt(784)) / 4

W = (40 ± 28) / 4

Now we have two possible solutions for the width, W:

1. W = (40 + 28) / 4 = 68 / 4 = 17

2. W = (40 - 28) / 4 = 12 / 4 = 3

Since the width cannot be less than the length, we discard the second solution (W = 3).

Now, substitute the value of W = 17 back into Equation 1 to find L:

2L + 2(17) = 40

2L + 34 = 40

2L = 40 - 34

2L = 6

L = 6 / 2 = 3

So, the dimensions of the pigpen are:
Length (L) = 3 feet
Width (W) = 17 feet

To find the dimensions of the pigpen, we need to use the given information about the fence and the area of the rectangular enclosure.

Let's assume the length of the rectangular pen is x feet. Since the enclosure is rectangular, the width will be 51 square feet divided by x feet.

Now, let's calculate the perimeter of the pigpen using the given 40 feet of fence:

Perimeter = 2*length + 2*width

Since the fence forms the perimeter of the rectangular pen, and we know the perimeter is 40 feet, we can set up the equation:

40 = 2*x + 2*(51/x)

Simplifying further:

40 = 2x + 102/x

Now, we can solve this equation to find the length (x) of the rectangular pen.

Multiply both sides by x:

40x = 2x^2 + 102

Rearrange the equation to form a quadratic equation:

2x^2 - 40x + 102 = 0

Now, we can solve this quadratic equation. Either by factoring, completing the square, or by using the quadratic formula.

Using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c:

x = (-(-40) ± sqrt((-40)^2 - 4(2)(102))) / (2(2))

Simplifying further:

x = (40 ± sqrt(1600 - 816)) / 4

x = (40 ± sqrt(784)) / 4

x = (40 ± 28) / 4

x = (40 + 28) / 4 or x = (40 - 28) / 4

x = 68 / 4 or x = 12 / 4

x = 17 or x = 3

Since it is not possible to have a width of 51/17 feet, we disregard the x = 17 solution.

Therefore, the dimensions of the pigpen are:

Length = 3 feet
Width = 51/3 = 17 feet

So, the pigpen has dimensions of 3 feet by 17 feet.