a farmer has enough fencing to build 40 feet of fence. He wishes to build a rectangular pen nest to his barn wall forming one side of the pen. What dimension should he make the pen so as to enclose the greatest possibility area?

length = x

width = y
area = A = x y
length of fencing = 2x+y = 40
so
y = 40-2x

A = x (40 - 2x)
A = 40 x - 2 x^2

maxmimum A , when dA/dx = 0
0 = 40 - 4 x
x = 10
then y = 40 - 20 = 20
area = A = 200

alternatively look at parabola
2 x^2 -40 x = -A
x^2 - 20 x = -(A/2)
x^2 - 20 x + 100 = -(A/2) + 100

(X-10)^2 = -(1/2)(A+200)
vertex at x = 10, A = 200

To find the dimensions that will enclose the greatest possible area, we can use the concept of calculus. Let's assume the width of the rectangular pen is "x" feet.

Given that the farmer has 40 feet of fencing, we can determine that the length of the pen (including the barn wall) will be 40 - x feet.

To calculate the area (A) of the pen, we multiply the length and width:

A = x * (40 - x)

To find the maximum area, we need to find the value of x that maximizes the function A(x).

To do this, we need to find the derivative of A with respect to x and set it equal to zero.

dA/dx = 40 - 2x

Setting 40 - 2x = 0 and solving for x, we get:

2x = 40
x = 20

So, the width of the pen should be 20 feet.

Since the length of the pen is given by 40 - x, the length would be:

40 - 20 = 20

Therefore, the farmer should make the pen 20 feet wide and 20 feet long to enclose the greatest possible area.

To find the dimensions that will enclose the greatest possible area with a given amount of fencing, we can use the concept of optimization. In this case, we can use the formula for the area of a rectangle, which is given by:

Area = Length × Width

Now, let's break down the information given in the question.

1. The farmer has enough fencing to build 40 feet of fence.
2. The barn wall will be one side of the pen.

To maximize the area, we need to consider how the fence is distributed. If we divide the 40 feet equally for the two widths and distribute the remaining length (which will be the length of the rectangle) along the length, it will create a square-shaped pen. However, to enclose the greatest possible area, we need to consider a rectangle shape.

Let's assume the length of the rectangle is x feet. Since we are considering the barn wall as one side, the remaining two sides will have a total length of (40 - x) feet.

Now, we can express the perimeter of the pen (which is equal to the total length of the two widths and the length of the rectangle) as:

Perimeter = 2 × Width + Length
Perimeter = 2 × (40 - x) + x

To find the value of x that maximizes the area, we need to differentiate the area formula with respect to x, set it equal to zero, and solve for x.

Area = Length × Width
Area = x × (40 - x)
Area = 40x - x^2

Differentiating the area with respect to x:
d(Area)/dx = 40 - 2x

Setting the derivative equal to zero to find the critical point:
40 - 2x = 0
2x = 40
x = 20

So, the critical point is x = 20, which means the length of the rectangular pen should be 20 feet.

To find the width, we can substitute this value of x into the equation for the perimeter:

Perimeter = 2 × (40 - x) + x
Perimeter = 2 × (40 - 20) + 20
Perimeter = 2 × 20 + 20
Perimeter = 60

Now, we know the width is 60 - 20 = 40 feet.

Therefore, to enclose the greatest possible area, the farmer should make the pen with dimensions 20 feet × 40 feet.

40=2L+w

w=40-2L

area= LW=L(40-2L)

Using calculus
dArea/dL=0=40-2L + l(-2)

4L=40
L=10 W= 20 so the pen is 10x20,with the barn serving as a 20ft side.

non calculus?
area=L(40-2L)
graph area vs L. Where is it maximum?