## Trending News

# Optimization Problem....?

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!

### 3 Answers

- 1 decade agoFavorite Answer
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!

- gp4rtsLv 71 decade ago
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.

- rajLv 71 decade ago
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