Question

The owner of a cattle ranch would like to fence in a rectangular area of 3000000 m^2 and then divide it in half with a fence down the middle. What is the minimum amount of fencing that the rancher must use?

Answer 1

If you notice the picture below, the amount of fencing, or perimeter, that will be used will be 3w + 2l

now    $\bf \begin{cases} A=l\cdot w\\\\\ A=3000000\\ ----------\\ 3000000=l\cdot w\\\\ \frac{3000000}{w}=l \end{cases}\qquad thus \\\\\\ P=3w+2l\implies P=3w+2\left( \cfrac{3000000}{w} \right)\implies P=3w+\cfrac{6000000}{w} \\\\\\ P=3w+6000000w^{-1}\\\\ -----------------------------\\\\ now\qquad \cfrac{dP}{dw}=3-\cfrac{6000000}{w^2}\implies \cfrac{dP}{dw}=\cfrac{3w^2-6000000}{w^2} \\\\\\ \textit{so the critical points are at } \begin{cases} 0=w^2\\\\ 0=\frac{3w^2-6000000}{w^2} \end{cases}$

solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum

notice the $0=w^2\implies 0=w$   so.  you can pretty much skip that one, though is a valid critical point, the width can't clearly be 0

so.. check the critical points on the other