A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 80ft of fence? 800 sq ft?

What should the dimensions of the garden be to give this area? 40ft is given so I answered with 40x20?

Is this correct?

Your answer is correct, but how did you set it up?

Did you use Calculus?

I am working with quadratic equations this week. Honestly, I drew a picture and made a logical guess.

x = length

z = width
2x + 2 z = 80
x+z = 40
Area = y = x z
so
y = x (40-x)
x^2 - 40 x = -y
that is a parabola, max or min at vertex (assume you do not do calculus but do know "completing the square", do so.
x^2 - 40 x + 20^2 = -y + 400
(x-20)^2 = - (y-400)
so, parabola sheds water (y small when x big positive or negative.
vertex - the maximum - at x = 20
then y = 20
(sure enough, a square)
at that vertex, y, the area is sure enough 400

vertex - the maximum - at x = 20

then z = 20
because x + z = 40

So I am wrong? It is 400 instead of 800 sq ft?

if it were 40 by 20, the perimeter would be:

40 + 20 + 40 + 20 = 120

But you only have 80 feet of fence.

always check.

20 + 40 + 20 + 40 = 120
You would have to stretch that 80 feet of fencing pretty hard.

I overlooked the barn being one side

perimeter = 2 x + z = 80
so z = 80 - 2 x

y = x z

y = x (80 - 2 x)

y = -2 x^2 + 80 x

2 x^2 - 80 x = -y

x^2 - 40 x = -y/2

x^2 - 40 x + 20^2 = -y/2 + 400

(x-20)^2 = - (y/2 - 400)

so x = 20
z = 80-40 = 40

area
y/2 = 400
y = area = 800

You guessed it right. I left out the wall of the barn.

LOL

Gives new meaning to the expression:

"Couldn't hit the side of a barn door...."

To determine if your answer is correct, let's go through the process of finding the dimensions that would yield the maximum area for the rectangular garden.

Let's assume the farmer uses the side of the barn as the length of the rectangle. We'll call this length "L". The remaining three sides will form the width of the rectangle, which we'll call "W".

The perimeter of a rectangle is given by the formula: P = 2L + 2W. In this case, we're given that the total fence length is 80ft, so we can set up the equation:

80 = 2L + 2W

Rearranging the equation:
2L = 80 - 2W
L = 40 - W

Now, let's consider the area of a rectangle which is calculated by multiplying the length (L) and width (W): A = L * W.

Substituting L with 40 - W, we have: A = (40 - W) * W.

To find the maximum area, we can use calculus by taking the derivative of A with respect to W and setting it equal to zero.

dA/dW = 40 - 2W

Setting dA/dW = 0 and solving for W:
40 - 2W = 0
2W = 40
W = 20

So, the width of the garden should be 20ft.

To find the length, we can use L = 40 - W:
L = 40 - 20
L = 20

Therefore, the dimensions of the garden to maximize the area with a total of 80ft of fence are 20ft by 20ft.

Now, let's check if the area corresponds to 800 sq ft:
A = L * W = 20 * 20 = 400 sq ft.

The area of the garden with dimensions 20ft by 20ft is 400 sq ft, which is not equal to 800 sq ft. Therefore, your initial answer of 40ft by 20ft is not correct.

The maximum area that the farmer can enclose with 80ft of fence is 400 sq ft, and the dimensions of the garden should be 20ft by 20ft.