# Optimization problem, fencing

• Apr 25th 2010, 09:07 PM
dorkymichelle
Optimization problem, fencing
11. A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one sides of the rectangle. how can he do this so as to minimize the cost of the fence.

i got this problem completely wrong, but here's what I did,
area = 1500000
area = xy = 1,500,000
length of fence = perimeter = 3x+4y
then
xy=1,500,000
y=\frac{1,500,000}{x}
3x+4\frac{1,500,000}{x}
f'(x) = 3+\frac{-6,000,000}{x^2}
= \frac{3x^2-6,000,000}{x^2}
= \frac{3(x^2-2,000,000)}{x^2}
x=1414
then... i got completely lost.
• Apr 25th 2010, 10:14 PM
Prove It
Quote:

Originally Posted by dorkymichelle
11. A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one sides of the rectangle. how can he do this so as to minimize the cost of the fence.

i got this problem completely wrong, but here's what I did,
area = 1500000
area = xy = 1,500,000
length of fence = perimeter = 3x+4y
then
xy=1,500,000
y=\frac{1,500,000}{x}
3x+4\frac{1,500,000}{x}
f'(x) = 3+\frac{-6,000,000}{x^2}
= \frac{3x^2-6,000,000}{x^2}
= \frac{3(x^2-2,000,000)}{x^2}
x=1414
then... i got completely lost.

Actually

$A = xy$ and $P = 3x + 2y$, since you only need one extra length to go down the middle. Since you want to minimise the cost of the fencing, you need to minimise the perimeter.

You know $A = 1.5\cdot 10^6 \,\textrm{ft}^2$.

So $xy = 1.5\cdot 10^6$

$y = \frac{1.5\cdot 10^6}{x}$.

Therefore

$P = 3x + 2y$

$= 3x + 2\left(\frac{1.5\cdot 10^6}{x}\right)$

$= 3x + \frac{3 \cdot 10^6}{x}$

$= 3x + 3\cdot 10^6 x^{-1}$.

$\frac{dP}{dx} = 3 - 3\cdot 10^6 x^{-2}$

$= 3 - \frac{3\cdot 10^6}{x^2}$.

This is minimised when the derivative is 0, so

$3 - \frac{3\cdot 10^6}{x^2} = 0$

$3x^2 - 3\cdot 10^6 = 0$

$x^2 - 10^6 = 0$

$x^2 = 10^6$

$x = 10^3$.

To be a minimum, the second derivative needs to be positive at that point.

$\frac{d^2P}{dx^2} = 6\cdot 10^6 x^{-3}$

$= \frac{6\cdot 10^6}{x^3}$

$= \frac{6\cdot 10^6}{(10^3)^3}$ when $x = 10^3$

$= \frac{6}{ 10^3}$

$> 0$.

So this is definitely the minimum.

When $x = 10^3$

$y= \frac{1.5\cdot 10^6}{10^3}$

$= 1.5\cdot 10^3$.

So the dimensions are:

$x = 10^3\,\textrm{ft}$ and $y = 1.5\cdot 10^3\,\textrm{ft}$, with $x$ as the length used to divide the paddock in half.