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EXAMPLE 3 Find the point on the parabola y - 4x that is closest to the point (2, 8). SOLUTION The distance between the point (2, 8) and the point (x, y) is d = - 2)² + (y – 8)² (See the graph.) But if (x, y) lies on the parabola, then x = y14, so the expression for d becomes d = + (y – 8)? (Alternatively, we could have substituted y = V4x to get d in terms of x alone.) Instead of minimizing d, we minimize its square: Video Example ) Tutorial d - f(y) = ( - 2)? + (y - 8)? Online Textbook (You should convince yourself that the minimum of d occurs at the same point as the minimum of d, but d is easier to work with.) Differentiating, we obtain f '(y) - 2(y/4 - 2)( ) + 2(y - 8) so f '(y) = 0 when y = Observe that f '(y) < 0 when y < 4 and f (y) > 0 when y > 4, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y = 4 (or we could simply say that because of the geometric nature of the problem, it's obvious that there is a closest point but not a furthest point). The corresponding value of x is x = y²/4 = . Thus the point on y = 4x closest to (2, 8) is (

EXAMPLE 3 Find the point on the parabola y - 4x that is closest to the point (2, 8). SOLUTION The distance between the point (2, 8) and the point (x, y) is d = - 2)² + (y – 8)² (See the graph.) But if (x, y) lies on the parabola, then x = y14, so the expression for d becomes d = + (y – 8)? (Alternatively, we could have substituted y = V4x to get d in terms of x alone.) Instead of minimizing d, we minimize its square: Video Example ) Tutorial d - f(y) = ( - 2)? + (y - 8)? Online Textbook (You should convince yourself that the minimum of d occurs at the same point as the minimum of d, but d is easier to work with.) Differentiating, we obtain f '(y) - 2(y/4 - 2)( ) + 2(y - 8) so f '(y) = 0 when y = Observe that f '(y) < 0 when y < 4 and f (y) > 0 when y > 4, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y = 4 (or we could simply say that because of the geometric nature of the problem, it's obvious that there is a closest point but not a furthest point). The corresponding value of x is x = y²/4 = . Thus the point on y = 4x closest to (2, 8) is (

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 3 Find the point on the parabola y = 4x that is closest to the point (2, 8). y SOLUTION The distance between the point (2, 8) and the point (x, y) is d = V (r – 2)² + (y - 8)2 2 6 10 (See the graph.) But if (x, y) lies on the parabola, then x = y/4, so the expression for d becomes d = + (y – 8)? (Alternatively, we could have substituted y = V4x to get d in terms of x alone.) Instead of minimizing d, we minimize its square: Video Example Tutorial d = f(y) = ( - 2)2 + (y - 8)2 Online Textbook (You should convince yourself that the minimum of d occurs at the same point as the minimum of d, but d is easier to work with.) Differentiating, we obtain f '(y) = 2(y/4 - 2)( ) + 2(y - 8) so f '(y) = 0 when y = Observe that f '(y) < 0 when y < 4 and f '(y) > 0 when y > 4, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y = 4 (or we could simply say that because of the geometric nature of the problem, it's obvious that there is a closest point but not a furthest point). The corresponding value of x is x = y/4 = . Thus the point on y = 4x closest to (2, 8) is (
Transcribed Image Text:EXAMPLE 3 Find the point on the parabola y = 4x that is closest to the point (2, 8). y SOLUTION The distance between the point (2, 8) and the point (x, y) is d = V (r – 2)² + (y - 8)2 2 6 10 (See the graph.) But if (x, y) lies on the parabola, then x = y/4, so the expression for d becomes d = + (y – 8)? (Alternatively, we could have substituted y = V4x to get d in terms of x alone.) Instead of minimizing d, we minimize its square: Video Example Tutorial d = f(y) = ( - 2)2 + (y - 8)2 Online Textbook (You should convince yourself that the minimum of d occurs at the same point as the minimum of d, but d is easier to work with.) Differentiating, we obtain f '(y) = 2(y/4 - 2)( ) + 2(y - 8) so f '(y) = 0 when y = Observe that f '(y) < 0 when y < 4 and f '(y) > 0 when y > 4, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y = 4 (or we could simply say that because of the geometric nature of the problem, it's obvious that there is a closest point but not a furthest point). The corresponding value of x is x = y/4 = . Thus the point on y = 4x closest to (2, 8) is (
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