<PRE><B>Question 889573</B>
Let w = width and L = length.
{{{L=2w-5}}} and {{{50&gt;2w+2L&gt;36}}}.


Wanting the longest length, first solve for w in terms of L.
{{{2w=L+5}}}
{{{w=(L+5)/2}}}, and substitute this into the inequality.


{{{50&gt;2(L+5)/2+2L&gt;36}}}
{{{50&gt;L+5+2L&gt;36}}}
{{{50&gt;3L+5&gt;36}}}
{{{50-5&gt;3L&gt;36-5}}}
{{{45&gt;3L&gt;29}}}
{{{29/3&lt;L&lt;45/3}}}
{{{highlight(29/3&lt;L&lt;15)}}}


For the LONGEST possible rectangle, L can be &lt;s&gt;&lt;s&gt;as near to 15 as desired but still must be less than&lt;/s&gt;&lt;/s&gt; 15.  A suitable value for w can then be found in order to agree with the inequality.
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