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Math - Questions and Answers from our Archives  Applied Problems
in Maxima and Minima  Date: 08/30/97 at 23:40:40 From: Anonymous
Subject: Applied problems in maxima and minima  tangent together
with the coordinate axes determine a triangle of  minimum area.
Answer: (2 (3)^(1/2) / 3, 8 / 3)  Please show me how to get the
answer.  Date: 08/31/97 at 12:38:45 From: Doctor Anthony
Subject: Re: Applied problems in maxima and minima  We can
simplify the working a little by considering the curve y = x^2,
and consider the triangle of minimum area formed by a tangent to
the  curve, the y axis, and the line y = 4.  This allows us to
use very  simple parametric coordinates (t,t^2) to represent the
curve.    dy/dx = (dy/dt)/(dx/dt) = 2t/1  =  2t  The equation of
the tangent is                        y-t^2 = 2t(x-t)  This
meets the y axis where x = 0, so                       y-t^2 =
-2t^2                          y = -t^2  as expected this is a
distance t^2 b