Calculus
posted by Cristy .
A line in the first quadrant is tangent to the graph of y=1/x^2. How can we minimize the area between the line and the axes?

Calculus 
Steve
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 (xa) + 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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