Posted by Marysvoice on Monday, February 18, 2008 at 9:34am.
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
So I am wrong? It is 400 instead of 800 sq ft?
always check.
20 + 40 + 20 + 40 = 120
You would have to stretch that 80 feet of fencing pretty hard.
vertex - the maximum - at x = 20
then z = 20
because x + z = 40
if it were 40 by 20, the perimeter would be:
40 + 20 + 40 + 20 = 120
But you only have 80 feet of fence.
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...."