Calculus word problem: thanks for any help!!?

Question

A farmer has 160 meters of fencing and he wants to build a rectangular enclosure for the chickens, using a hedge as one side of the rectangle. What is the maximum area of such an enclosure?

Answer 1

Onyl one 'y' because the other side is the hedge.

2x + y = 160
xy = A This is what we want to maximize.

y = -2x + 160

Substitute:
x(-2x + 160)
-2x^2 + 160x

Derivitive:
0 = -4x + 160
-4x = -160
x = -160/-4
x = 40

The maximum area is when x = 40
2(40) + y = 160
y = 80

(80 feet of hedge)

2(40) + 80 = 160ft
(40) (80) = 3200 sq ft

Answer 2

Hello,

Let w = width
Let l = length so

perimeter = 160 = l + 2w
area = lw Solve the first one for l giving us l = 2w - 160 now put this into the second one and we have

area = (2w - 160)w = 2w^2 -160w take the derivative giving us

a' = 4w - 160 set = 0 0 = 4w - 160 and the w = 40 meters so put it back into the first on 160 = l + 2*40 so l = 80 meters.

Hope This Help!!

Answer 3

Let w be the width. Thus 160-2w is the length.
area = f(w) = w(160-2w)
f'(w) = 160 - 2w - 2w = 0 => w = 40
f(40) = 40(160-80) = 3200 m^2, the maximum area