1
$\begingroup$

Another question! Very excited to be joined to this community. I highly appreciate the help and intelligence around here.

Basically, a guy named Ron is trying to decide whether a rectangular or circular section will produce a larger area when creating a fence using 100 metres of barricade mesh by working out and comparing the two. I understand how to find the maximum areas of both the rectangular and circular areas.

However, the second part of the question now asks for me to find the maximum area of both a rectangular and circular section using n metres of barricade mesh. In my head I believe this is quite simple but I'm getting confused with substitution and such. For example, for the circular section we know that n is the circumference. I wanted to find the radius of this and then work out the area that way, but the radius is r = C/2π, which would make the equation r = n/2π.

And then putting that into the equation to find the area of a circle uses Area = πr^2, and then it gets very confusing because it becomes Area = π(n/2π)^2. I'm just completely confused after that.

The rectangle area is also confusing me, as I'm doing this in a step-by-step order.

For example, n = 2l + 2w (l = length and w = width). It also means that n/2 = l +w.

w in terms of l is equal to w = n/2 + l.

Finding the maximum area uses l(n/2 + l), which I have gotten very stuck on and have worked it out as ln/2 + l^2.

Furthermore, while this could be right, how am I meant to find the length used to find the maximum area? I've found that it is always a quarter of the starting length given. Then this all becomes very confusing as l = n/4 and I have no clue on how to substitute that into ln/2+l^2.

Please excuse my messy explanation. I don't even know if this makes sense! TLDR; I'm trying to find the maximum area of both a rectangular and circular sectioning (separately) using n metres as the "maximum" length to work with. Any help is appreciated!

EDIT: Seems as though I've gotten this right! One thing I did get wrong, however, is ln/2+l^2. My previous calculation forgot to make l^2 negative and thus resulted in a much larger result. It should have been ln/2–l^2 and then substitute l as n/4.

  • 0
    Your computation for the area of the circle with circumference $n$ is correct. It simplifies as $A=\pi(\frac{n}{2\pi})^2=\frac{n^2}{4\pi}$. – 2017-01-27
  • 0
    Thank you for that clarification! I was getting quite confused there. – 2017-01-27

1 Answers 1

0

Since you understand the circle part of the problem where the maximum area is $A=\dfrac{n^2}{4\pi}$ I will discuss the rectangle. By the way you wrote in your question that $W=\dfrac{n}{2}+L$ but it is $W=\dfrac{n}{2}-L$.

The largest rectangle from $n$ meters of fencing is actually the square with $L=W=\dfrac{n}{4}$ which gives an area $A=\dfrac{n^2}{16}$.

To see why this is so without using calculus suppose we try to make the length slightly larger than $\dfrac{n}{4}$ by letting $L^*=\dfrac{n}{4}+E$. Then the width will have to be $W^*=\dfrac{n}{4}-E$ to keep the perimeter equal to $n$. But now the area is $A^*=\left(\dfrac{n}{4}+E\right)\cdot\left(\dfrac{n}{4}-E\right)=\dfrac{n^2}{16}-E^2$ which is smaller than the area $A=\dfrac{n^2}{16}$ when $L=W=\dfrac{n}{4}$.

If we try to make the length slightly smaller, say $\dfrac{n}{4}-E$ instead of slightly larger then the width will have to be $\dfrac{n}{4}+E$ and so we still get an area $\dfrac{n^2}{16}-E^2$ which is smaller.

Since $\dfrac{n^2}{4\pi}>\dfrac{n^2}{16}$ the circle wins.