<PRE><B>Question 1180422</B>
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The formal solution from tutor @ikleyn is fine.  It is a good example of problem solving using formal algebra; you should understand it.&lt;br&gt;
The solution method shown by the other tutor is NOT irrelevant, as tutor @ikleyn says.  It is a fast path to the solution -- &lt;b&gt;&lt;i&gt;IF&lt;/b&gt;&lt;/i&gt; you use it correctly, which the other tutor did not.&lt;br&gt;
It is relatively easy to prove that in any problem like this, the maximum area is obtained when -- as the other tutor says -- the total amount of fencing is equally divided between the lengths and widths.&lt;br&gt;
So the initial starting point of the other tutor&#39;s response was fine; she just didn&#39;t use the right numbers.&lt;br&gt;
So while the formal algebraic solution is fine if a formal solution is required, the problem can be solved easily and quickly if an informal solution is allowable.&lt;br&gt;
The total length of fencing is 40m, so we want to use 20m for the length(s) and 20m for the width(s).&lt;br&gt;
In this problem, there is one length of fencing and 5 widths of fencing, so the maximum area is when the length is 20/1=20m and the width is 20/5=4m.&lt;br&gt;
ANSWER: The maximum area is 20m*4m = 80 square meters&lt;br&gt;
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