Calculus

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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 -

    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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