Posted by Cristy on Wednesday, August 1, 2012 at 10:00pm.
the line touching the graph at x=a has slope -2/a^3
The line through (a,1/a^2) with slope -2/a^3 is
y = -2/a^3 (x-a) + 1/a^2
This line crosses the axes at (0,3/a^2) and (3a/2,0)
So, the area of the triangle formed by the line and the axes is
A(a) = 1/2 * 3/a^2 * 3a/2 = 9/(4a)
Now we have us a problem. dA/da is never 0. The area is infinite at a=0, and goes to zero at a = oo
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