Optimization Problem....?

Question

A garden is to be laid out in a rectangular area and protected by a chicken wire fence. What is the larges possible area of the garden if only 100 running feet of chicken wire is available for the fence?
Thanks for any help!

Answer 1

let length = x
breadth = 50-x
area = x*(50-x)
differntiate it with respect to x and equate it to zero!
50-2x = 0 that is x=25; 50-x = 25
so the largest area is 25*25=625 square feet!

Answer 2

The fence forms the perimeter of the garden. The perimeter is P = 2L+2W, where L and W are the length and width of the garden. The area is A =LW. Solve for L or W from the perimeter equation: for example L = .5*(P - 2W); then plug that in for L in the area equation to get an equation in W:

A = .5*(P- 2W)*W
A = .5*PW - W^2

Take the derivative of A to get A' = .5*P - 2W
set this equal to 0 and sovle for W. You will get w = P/4. From the perimeter equation you will get L = P/4. In other words, the gardent is a square with 1/4 of the fence length forming each side.

Answer 3

material needed=2pir(h+r)
volume=pir^2h
pir^2h=1000cc (given)
h=1000/(pir^2)
substituing h in the equation 2pir(h+r)
material=2pir(1000/pir^2+r)
=2pir*1000/pir^2+2pir^2
dA/dt=-2000/r^2+4pir
setting this to zero
-2000+4pir^3=0
4pir^3=2000
r^3=2000/4pi
r=cube root of 500/3.14
=5.4 approx
substituting
h=1000/(3.14*5.4*5.4)
=10.9 aapprox
so the radius is 5.4 cm approx
and the height 10.9 approx