# Math

posted by on .

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?

• Math - ,

Did you use Calculus?

• Math - ,

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

• Math - ,

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

• Math - ,

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

• Math - ,

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

• Math typo - ,

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

• Math - ,

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

40 + 20 + 40 + 20 = 120

But you only have 80 feet of fence.

• Whoa - sorry, missed the barn - ,

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.

• Whoa - sorry, missed the barn - ,

LOL
Gives new meaning to the expression:

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