Question

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

Short answer

He could choose the dimension 1000 ft by 1500 ft to minimize the cost of the fence.

Step-by-step solution

Optimization of a Function

The Optimization of a Function \(f\left( x \right)\) for any real value of \(c\) can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\).

Establishing the relations using the data given in the question.

The given area of the rectangular field is: \(1.5 \times {10^6}{\rm{ sq}}{\rm{.feet}}\)

Let the whole scenario can be designated as shown:

where,

\(\begin{aligned}{l}y \to {\rm{length}}\\x \to {\rm{breadth}}\end{aligned}\)

So, the area of the field will be:

\(\begin{aligned}{c}xy = 1.5 \times {10^6}\\ \Rightarrow y = \frac{{1.5 \times {{10}^6}}}{x}\end{aligned}\)

From the figure, the overall length of fencing will be:

\(\begin{aligned}{c}L = 3x + 2y\\L\left( x \right) = 3x + 2\left( {\frac{{1.5 \times {{10}^6}}}{x}} \right)\\ = 3x + \frac{{3 \times {{10}^6}}}{x}\end{aligned}\)

Optimizing the function according to the given condition

For minimization, we have:

\(\begin{aligned}{c}L'\left( x \right) = \frac{d}{{dx}}\left( {3x + \frac{{3 \times {{10}^6}}}{x}} \right)\\\frac{d}{{dx}}\left( {3x + \frac{{3 \times {{10}^6}}}{x}} \right) = 0\\3 - \frac{{3 \times {{10}^6}}}{{{x^2}}} = 0\\{x^2} = {10^6}\\x = {10^3}\end{aligned}\)

Now, we have:

\(\begin{aligned}{l}{\rm{for }}0 < x < {10^3} \to L'\left( x \right) < 0\\{\rm{and}}\\{\rm{for x}} > {10^3} \to L'\left( x \right) > 0\end{aligned}\)

Therefore, the function will have an absolute minimum value at \(x = 1000{\rm{ ft}}\). For this, we have:

\(\begin{aligned}{c}y = \frac{{1.5 \times {{10}^6}}}{x}\\ = \frac{{1500000}}{{1000}}\\ = 1500{\rm{ ft}}\end{aligned}\)

Hence, the cost of fencing will be minimum for dimensions \(1000{\rm{ ft and }}1500{\rm{ ft}}\).