<PRE><B>Question 811291</B>
L + 2W = 100&lt;P&gt;
A = LW&lt;P&gt;
Find, from the first equation, one variable in terms of the other.  L = 100-2W.  Substitute that into the area equation.&lt;P&gt;
A = (100-2W)W = {{{100W - 2W^2)&lt;P&gt;
Use the first derivative to find the maximum possible area.&lt;P&gt;
A&#39; = 100 - 4W = 0&lt;P&gt;
4W = 100 so W = 25&lt;P&gt;
With these problems the value found when setting the derivative = to 0 us the max or min.  But you can verify using the first derivative test.&lt;P&gt;
0 &lt;= W &lt;= 50, so the ranges to check are 0 to 50 and 50 t0 100&lt;P&gt;
W=1 is in the first range.  Plug into the first derivative.  100- 4 = 96 which is positive.&lt;P&gt;
w=50 is in the second range.  100-4(50) is negative.&lt;P&gt;
The first derivative is positive before W=25 and negative after, so W=25 is a max.&lt;P&gt;
L + 2(25) = 100 so L = 100-50 = 50.&lt;P&gt;</PRE>