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October 2, 2014

October 2, 2014

Posted by **Cristy** on Wednesday, August 1, 2012 at 10:00pm.

- Calculus -
**Steve**, Thursday, August 2, 2012 at 5:30amthe 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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