How to find the maximum perimeter of a rectangular surface when one of its sides must be an odd length?

Jun 2017
Lima, Peru
The problem is as follows:

A radio technician is tasked to accommodate 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:

1.&\textrm{52 meters}\\
2.&\textrm{36 meters}\\
3.&\textrm{48 meters}\\
4.&\textrm{44 meters}\\

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. Help please!


Math Team
Sep 2015
$l = w + 4\\
w = 2k+1,~k \in \mathbb{Z^+}\\
l w \leq 165\\
(w+4)(2k+1) \leq 165\\
(2k+1+4)(2k+1) \leq 165\\
(2k+5)(2k+1) \leq 165\\
4k^2 +12k + 5 \leq 165\\
4k^2 + 12k -160 \leq 0\\
k^2 +3k -40 \leq 0\\
(k+8)(k-5) \leq 0\\
k = 5\\
w = 11\\
l = 15\\
p = 2(11+15) = 52
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