Math
posted by Marysvoice 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?

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 (40x)
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
(x20)^2 =  (y400)
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
(x20)^2 =  (y/2  400)
so x = 20
z = 8040 = 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...."