Explanation Given information: The equation of the line is as follows. x a + y b = 1 (1) Calculation: Rearrange Equation (1), y b = 1 − x a y = b − b x a (2) The distance from origin to the line at any point is as follows. D = ( x − 0 ) 2 + ( y − 0 ) 2 = x 2 + y 2 (3) Substitute ( b − b x a ) for y in Equation (3). D = x 2 + ( b − b x a ) 2 = x 2 + b 2 − 2 b 2 x a + b 2 x 2 a 2 (4) Differentiate both sides of the Equation (4) with respect to x . D ′ ( x ) = 2 x + 0 − 2 b 2 a + 2 b 2 x a 2 x 2 + b 2 − 2 b 2 x a + b 2 x 2 a 2 = 2 x − 2 b 2 a + 2 b 2 x a 2 x 2 + b 2 − 2 b 2 x a + b 2 x 2 a 2 = 2 ( x − b 2 a + b 2 x a ) 2 x 2 + b 2 − 2 b 2 x a + b 2 x 2 a 2 = x − b 2 a + b 2 x a x 2 + b 2 − 2 b 2 x a + b 2 x 2 a 2 Substitute 0 for D ′ ( x )

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 4.6, Problem 28E

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Find the point on the line xa+yb=1.

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Chapter 4.6, Problem 27E

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Chapter 4 Solutions

University Calculus: Early Transcendentals (4th Edition)

Ch. 4.1 - In Exercises 16, determine from the graph whether...Ch. 4.1 - In Exercises 1–6, determine from the graph whether...Ch. 4.1 - Prob. 3E Ch. 4.1 - In Exercises 1–6, determine from the graph whether...Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - In Exercises 7–10, find the absolute extreme...Ch. 4.1 - In Exercises 7–10, find the absolute extreme...Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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Find the point on the line x a + y b = 1 .

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