Question

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?

Answer 1

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.

Answer 2

LOL
Gives new meaning to the expression:

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

Answer 3

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

Answer 4

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

Answer 5

vertex - the maximum - at x = 20
then z = 20
because x + z = 40

Answer 6

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

Answer 7

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

Answer 8

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

40 + 20 + 40 + 20 = 120

But you only have 80 feet of fence.

Answer 9

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