# Optimization: A farmer wants to fence an area of 1.5 million square feet in a rectangular field?

Optimization: A farmer wants to fence an area of 1.5 million square feet in a rectangular field?

• For the sake of argument, letâ€™s say that “length” is the dimension that the middle partition is parallel to, and the other two sides are the “width”. Let L be the length and W be the width. Then the area is:

LW = 1.5 x 10^6 square feet

The total amount of fencing used, in feet, is going to be the sum of the four sides, plus the partition that runs down the middle:

(L+L+W+W) + L =

3L + 2W

Using the first equation to substitute, you get

3L + [2(1.5 x 10^6)/L]

Now that you have the fencing equation in terms of just L, you can take the derivative, set it equal to 0, then solve for L. This tells you the value of L that minimizes the amount of fencing used, which in turn means that it minimizes the cost. To find W, thatâ€™s just W = (1.5 x 10^6 square feet)/L.

• Let the dimensions of the rectangle be x and y, so the length of fencing required is given by:

L = 3x + 2y

And the area, A is given by

A = xy = 1500000 â€“> y = 1500000/x

so L = 3x + 3000000/x

dL/dx = 3 â€“ 3000000/x^2 = 0 (at maximus)

x^2 = 1000000

x = 1000 â€“> y = 1500

Source(s): I know stuff.
• p = width of the field

q = length

pq = 1.5 x 10^6 ftÂ²

Perimeter to be minimized = 2p + 3q

P = 2p + 3(1.5 x 10^6 / p)

P = 2p + (4.5 x 10^6)p^-1

dP/dp = 2 + (4.5 x 10^6)(-1)p^-2

dP/dp = 2 â€“ (4.5 x 10^6) / pÂ²

Set dP/dp = 0 to find the minimumâ€¦

2 â€“ 4.5 x 10^6 / pÂ² = 0

2pÂ² â€“ 4.5 x 10^6 = 0

2pÂ² = 4.5 x 10^6

pÂ² = 9 x 10^6

p = âˆš(9 x 10^6)

p = 3000 ft <===================

q = (1.5 x 10^6) / 3000 = 500 ft <==============

• enable container have sides L and B and enable fence be parallel to section L Perimeter + divider = 3L + 2Bâ€¦â€¦â€¦â€¦â€¦â€¦..(a million) element of container = L*B = 3000000 So, B = 3000000/L replace this value of B in (a million) Fencing = 3L + (6000000/L) or (3L^2 + 6000000) / L