Question

It is required to inclose a rectangular field by a fence and then to divide it into two lots by a fence parallel to one of the sides. If the area of the field is given, find the ratio of the sides so that the total length of fence shall be a minimum.

Answer 1

Let $x=$ the length of the field, let $y=$ the width of the field.

Given $A=xy=const.$

Then $y=\dfrac{A}{x}.$

The total length $P$ of fence will be

$$P=2x+2y+y$$

Substitute

$$P=P(x)=2x+3(\dfrac{A}{x})$$

Differentiate with respect to $x$

$$P'=2-\dfrac{3A}{x^2}, x>0$$

Find the critical number(s)

$$P'=0=>2-\dfrac{3A}{x^2}=0$$

$$x=\pm \sqrt{\dfrac{3A}{2}}$$

Since $x>0$ we take $x=\sqrt{\dfrac{3A}{2}}.$

If $0<x<\sqrt{\dfrac{3A}{2}}, P'<0, P(x)$ decreases.

If $x>\sqrt{\dfrac{3A}{2}}, P'>0,P(x)$ increases.

The function $P(x)$ has a local minimum at $x=\sqrt{\dfrac{3A}{2}}.$

Since the function $P(x)$ has the only extremum for $x>0,$ then the function $P(x)$ has the absolute minimum for $x>0$ at $x=\sqrt{\dfrac{3A}{2}}.$

$$y=\dfrac{A}{\sqrt{\dfrac{3A}{2}}}=\sqrt{\dfrac{2A}{3}}$$

$$length/width=\dfrac{x}{y}=\dfrac{\sqrt{\dfrac{3A}{2}}}{\sqrt{\dfrac{2A}{3}}}=\dfrac{3}{2}$$

if the fence is parallel to the width of the field then the ratio $length/width$ is $3/2.$