<PRE><B>Question 1141515</B>
&lt;br&gt;
Let x be the angle the sides make with a perpendicular to the bases -- e.g., if the angles the sides make with the bottom base are 100 degrees, then x is 10 degrees.  Then&lt;br&gt;
the bottom base is 10
the top base is 10+2*10sin(x)
the height (depth) is 10cos(x)&lt;br&gt;
The maximum amount of water is when the cross sectional area is maximum.&lt;br&gt;
The area is height times the average of the two bases.&lt;br&gt;
{{{A = 10cos(x)((10+(10+20sin(x)))/2)}}}&lt;br&gt;
{{{A = 100cos(x)(1+sin(x))}}}&lt;br&gt;
To find the maximum area, find where the derivative is zero.  Note we can ignore the constant 100 and find the maximum value of {{{cos(x)(1+sin(x))}}}&lt;br&gt;
By the product rule, the derivative is...&lt;br&gt;
{{{(cos(x)*cos(x))+(-sin(x))(1+sin(x))}}}
{{{cos(x)^2-sin(x)-sin(x)^2}}}&lt;br&gt;
Using {{{cos(x)^2-sin(x)^2 = cos(2x),&lt;br&gt;
{{{cos(2x)-sin(x)}}}&lt;br&gt;
The derivative is zero when cos(2x) = sin(x).&lt;br&gt;
Knowing that the angle x is acute, and that for acute angles sin(x) = cos(90-x), we can determine that the derivative is zero when x is 30 degrees.&lt;br&gt;
This can be confirmed by graphing the area function on a graphing calculator and finding the value of x that produces the maximum area.&lt;br&gt;
ANSWER: The gutter will hold the maximum amount of water when the length of the top base is 10+20*sin(30) = 10+20(1/2) = 10+10 = 20.
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