Calculus Questions...?

Question

A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 180,000 sq meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river.

Answer 1

p = 2x + y
a = xy
sub on equation into the other
y = a/x
p = 2x + a/x

dy/dx = -a/x^2 + 2 = 0

Answer 2

Use Area of a rectangle = Lw
A = Lw
A = 180 000
w= (180 000)/ (L)
Perimeter = P = 2L + 2w
Note that we only need 3 sides so the formula actually becomes
P = 2L + w
Substitute for w and
P = 2L + (180 000)/L
Use the power rule to find P'
P' = 2 - (180 000)/ L^2
Set the P' = 0
(180 000)/ L^2 = 2
180 000 = 2L^2
L^2 = 90 000
L = 300
Plug back in for w
w = (180 0000 / (300)
w = 600
The dimensions for the least amount of fencing is 300m by 600 m

Answer 3

Let one side be x and the side with the river be y. The area is

A = 180,000 = xy

The perimeter of the fence is

P = y + 2x

Using area

y =A/x so that P = A/x +2x

Now if P is to be a minimum, dP/dx = 0

dP/dx = 0 = -A/x^2 + 2 ---> 2x^2 = A

x = sqrt(A/2) = sqrt(90,000) = 300 m

Then y = A/x = 180000/300 = 600 m