Applied Problems in Maxima and Minima
Date: 08/30/97 at 23:40:40 From: Anonymous Subject: Applied problems in maxima and minima At what first quadrant point on the parabola y = 4 - x^2 does the 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 below the x axis. The vertical side of the required triangle will therefore be of length 4+t^2. The tangent meets y = 4 where 4-t^2 = 2tx-2t^2 4 + t^2 = 2tx x = (4+t^2)/2t Area of triangle = (1/2)(4+t^2)(4+t^2)/2t A = (1/4t)(16 + 8t^2 + t^4) = (1/4)(16/t + 8t + t^3) dA/dt = (1/4)(-16/t^2 + 8 + 3t^2) = 0 for max or min. So 3t^2 + 8 = 16/t^2 3t^4 + 8t^2 - 16 = 0 (3t^2-4)(t^2+4) = 0 and so t^2 = 4/3 t = 2/sqrt(3) Thus the x coordinate is 2/sqrt(3). This is the same x value as for the equation y = 4 - x^2, and therefore we can substitute x = 2/sqrt(3) in this equation to find the y coordinate. y = 4 - 4/3 = 8/3 Coordinates of required point [2/sqrt(3), 8/3] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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