Posted by **Marysvoice** on Monday, February 18, 2008 at 9:34am.

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 -
**Reiny**, Monday, February 18, 2008 at 9:39am
Your answer is correct, but how did you set it up?

Did you use Calculus?

- Math -
**Damon**, Monday, February 18, 2008 at 9:59am
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 typo -
**Damon**, Monday, February 18, 2008 at 10:06am
vertex - the maximum - at x = 20

then z = 20

because x + z = 40

- Math -
**Damon**, Monday, February 18, 2008 at 10:15am
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 -
**Damon**, Monday, February 18, 2008 at 10:40am
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.

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