Question 1152184
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Let the isosceles triangle have the center of the base at (0,0); let the vertices be at (-b,0), (0,a), and (b,0).  For help with the problem later on, draw the altitude from (0,0) to (0,a).<br>
Let one vertex of the rectangle be at (x,0), with 0 < x < b.  Draw that rectangle.<br>
The equation of the "right" side of the triangle, with endpoints at (0,a) and (b,0), is<br>
{{{y = (-a/b)x+a}}}<br>
Then the dimensions of the rectangle are 2x and (-a/b)x+a.  And the area is then<br>
{{{A = (2x)((-a/b)x+a) = (-2a/b)x^2+2ax}}}<br>
To find the maximum area, find the condition that makes the derivative of the area function equal to 0:<br>
{{{dA/dx = (-4a/b)x+2a}}}<br>
{{{(-4a/b)x+2a = 0}}}
{{{2a = (4a/b)x}}}
{{{x = b/2}}}<br>
So the maximum area of the rectangle is obtained when one vertex is halfway between the center of the base and one endpoint of the base.<br>
When a sketch is made of that condition, it should be obvious that the area of the rectangle is equal to the area of the four small triangles; and that means the area of the rectangle is half the area of the original triangle.<br>