<PRE><B>Question 1150998</B>
.


&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The correct solution is below.



&lt;pre&gt;
You determine first the area of the class room for 22 students

    Area = 25 + 22*1.5 = 58 square meters.


A rectangle which provides the minimum perimeter at given area, is THE SQUARE with the side length equal to square root of the area.


In this case, the classroom with minimal perimeter is the square with the side length of

    {{{sqrt(58)}}} = 7.616 meters.    


The least possible perimeter is then  4*7.616 = 30.464 meters (approximately).       &lt;U&gt;ANSWER&lt;/U&gt;
&lt;/pre&gt;

Solved.


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If you want to see many other similar optimization problems, &amp;nbsp;look into my lesson

&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;- &lt;A HREF=https://www.algebra.com/algebra/homework/word/misc/Calculus-optimization-problems.lesson&gt;Calculus optimization problems&lt;/A&gt;

in this site.


Specifically, &amp;nbsp;for the given problem, &amp;nbsp;its precise analogue is the &amp;nbsp;&lt;U&gt;Problem 3&lt;/U&gt; &amp;nbsp;of the referred lesson.



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