Question

A farmer plans to fence a rectangular pasture adjacent to & river (see the figure below):

The pasture must contain $720,000$ square meters in order to provide enough grass for the herd.

No fencing is needed along the river .

What dimensions will require the least amount of fencing?

Answer 1

Finding the dimensions which will require the least amount of fencing:

Step-1: Finding the expression for width

The given area is: $A=720,000\text{sq.mtrs}$

Let us assume that,

$$\begin{gathered} \text{⇒length=l} \\ \text{⇒width=w} \end{gathered}$$

Area of the rectangle can be expressed as,

$A=l\times w$

Substitute $A=720,000\text{sq.mtrs}$in the above Equation.

$$\begin{gathered} 720,000=l\times w \\ w=\frac{720,000}{l}......\left(1\right) \end{gathered}$$

Step-2: Finding expression for perimeter

Formula for the perimeter can be expressed as,

$p=2\left(l+w\right)$

Rewrite the above Equation as,

$p=2l+2w$

Because one side is along the river. then substitute $w=\frac{720,000}{l}$ in the above Equation.

$p=2l+2\times \frac{720,000}{l}$

Step-3: Finding maxima and minima for perimeter value

Differentiate the above Equation with respect to $l$

$$\begin{gathered} \frac{dp}{dl}=2-2\times \frac{720,000}{l^2} \\ \frac{d^{2}p}{d{l}^2}=-720,000\frac{-4}{l^3} \\ \frac{d^{2}p}{d{l}^2}=\frac{2880,000}{l^3} \end{gathered}$$

Find the vale of $l$ and $w$

$$\begin{gathered} \Rightarrow \frac{dp}{dl}=2-\frac{1440,000}{l^2} \\ \Rightarrow 2=\frac{1440,000}{l^2}\left[\frac{dp}{dl}=0\right] \\ \Rightarrow l^2=720,000 \\ \Rightarrow l=\sqrt{720,000} \\ \Rightarrow l=600\sqrt{2}\text{m} \end{gathered}$$

Substitute $l=600\sqrt{2}\text{m}$ is a minimum point in Equation (1)

$$\begin{gathered} w=\frac{720,000}{600\sqrt{2}} \\ w=600\sqrt{2}\text{m} \end{gathered}$$

Step-4: Finding value of minimum perimeter

So minimum perimeter can be expressed as,

$$\begin{gathered} p_{min}=2l+2w \\ {p}_{min}=2\left(600\sqrt{2}\text{m}\right)+2\left(600\sqrt{2}\text{m}\right) \\ {p}_{min}=2400\sqrt{2}\text{m} \end{gathered}$$

Hence, the dimensions will require the least amount of fencing is $\text{2400}\sqrt{2}\text{m}$.