The problem is as follows:

A radio technician is tasked to accomodate an antenna and other supplies for a tv broadcast in a certain terrain of rectangular shape. This terrain must have a large which exceeds its width in $4$ meters. The area is required to be less than $165$ square meters and its width must have an odd length also measured in meters. Given these conditions find maximum perimeter of such terrain.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{52 meters}\\ 2.&\textrm{36 meters}\\ 3.&\textrm{48 meters}\\ 4.&\textrm{44 meters}\\ \end{array}$

How exactly can I find the maximum perimeter?. What it came to my mind was to attempt using the first derivative as it is mentioned that:

$\textrm{x = width of the terrain}$

$\textrm{x+4 = large of the terrain}$

Therefore the area is:




Hence solving this results into:


But this doesn't yield exactly a value which I could use:

Then I attempted to do this by trying the first derivative as:



Then this would be the minimum:


But this is not what I'm being requested.

Even if I attempt to evaluate the function it does yield:


which is negative and it doesn't make sense.

What would be the right approach for this problem?. Can someone help me and indicate which part did I made a mistake or conceptual flaw?. It would help a lot to include some step by step solution to see what I did wrong.

  • 1
    $\begingroup$ What does "large of the terrain" mean? Did you intend "large" to be "length" instead? $\endgroup$ Mar 21 '20 at 22:45
  • $\begingroup$ @JohnOmielan The intended meaning was the length of the terrain, whose side is larger. Is it understood better?. In other words $(l\times w)$. $\endgroup$ Mar 21 '20 at 23:58
  • $\begingroup$ @JohnOmielan The existing answer indicates to consider the maximum perimeter to be less and equal but in the given condition of the problem states to be only less than $165$ square meters. Wouldn't it make the maximum perimeter to be $44$ meters? or did I made any flaw on it?. $\endgroup$ Mar 21 '20 at 23:59
  • $\begingroup$ Thanks for the clarification of what you intended. However, to me at least, it would have been clearer to write "length" instead, as that is always the opposite dimension of width in $2$ dimensions. Also, the length is usually considered larger than the width, that is how which dimension to label length and which to label width. Nonetheless, you've made this explicit by stating the length must be larger than the width by $4$ meters. As for your result & that of the given answer, note the answer has now been corrected. $\endgroup$ Mar 22 '20 at 0:16

Going back to your quadratic, just use the positive solution. That's the maximum value for $x$. The perimeter is $4x+8$, increasing with $x$. So the maximum perimeter is at the largest $x$. In addition, they tell you that $x$ is odd. Choose the largest odd number, less then $-2+13=11$. Therefore $x=9$ and the perimeter is $44$

  • $\begingroup$ The reason why I stuck at the quadratic was due the value of the square root I mean $\sqrt{676}$. This result isn't nice. Yes I noticed that x has to be the largest. So do you mean that to solve this, what's only needed is to try out all odds starting from $1$? $\endgroup$ Mar 21 '20 at 23:44
  • 1
    $\begingroup$ $\sqrt{676}=26$ $\endgroup$
    – Andrei
    Mar 21 '20 at 23:55
  • $\begingroup$ In this question it was indicated that the area of the terrain to be only less than $165$ square meters. Wouldn't it make that way to be $44\,m$ the maximum perimeter?. Does this problem requires or could be also solved using derivatives?. Hence this part has left unattended in my question. $\endgroup$ Mar 22 '20 at 0:01
  • $\begingroup$ Sorry I'm doing the square roots by hand. I noticed that I made a mistake, but one answer is $x<15$ or $x<-11$ so I'm selecting the odd for the $x<15$ hence it can be $13$ but it doesn't check the other choice can be $11$ but it makes it equal to $165$ and the problem states it must be less than $165$ hence it has to be $9$, therefore. The maximum perimeter must be $44\,m$. Am I right with this? $\endgroup$ Mar 22 '20 at 0:07
  • $\begingroup$ Yes. The approach as you described works for any numbers, even if the numbers are not nice. You are also right about the number: $x=11$ will give an area exactly 165. So you need to choose $x=9$. Then the perimeter is $4*9+8=44$ $\endgroup$
    – Andrei
    Mar 22 '20 at 0:13

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