Explanation Given: f ( c , d ) = ( c − a 1 ) 2 + ( d − b 1 ) 2 + .... + ( c − a n ) 2 + ( d − b n ) 2 Let f ( x , y ) be a continuous function on a closed, bounded domain D in R 2 . Then: (i) f ( x , y ) takes on both a minimum and a maximum value on D . (ii) The extreme values occur either at critical points in the interior of D or at points on the boundary of D . Calculation: f ( c , d ) = ( c − a 1 ) 2 + ( d − b 1 ) 2 + .... + ( c − a n ) 2 + ( d − b n ) 2 f c ( c , d ) = 2 ( c − a 1 ) + ... + 2 ( c − a n ) f d ( c , d ) = 2 ( d − b 1 ) + ... + 2 ( d − b n ) For the critical points, computing f c ( c , d ) = 0 and f d ( c , d ) = 0 , we get 2 ( c − a 1 ) + ... + 2 ( c − a n ) = 0 ( c − a 1 ) + ... + ( c − a n ) = 0 n c − ( a 1 + ... + a n ) = 0 c = a 1 + ... + a n n 2 ( d − b 1 ) + ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 15.7, Problem 49E

To determine

We need to show that the sum of the square of the distances from a point $P = (c, d)$ to $n$ fixed points $(a_{1}, b_{1}), ... (a_{n}, b_{n})$ is minimized, when $c$ is the average of the $x$ -coordinates $a_{i}$ and $d$ is the average of the $y$ -coordinates.

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Chapter 15 Solutions

Calculus - Standalone book

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E

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We need to show that the sum of the square of the distances from a point P = ( c , d ) to n fixed points ( a 1 , b 1 ) , ... ( a n , b n ) is minimized, when c is the average of the x -coordinates a i and d is the average of the y -coordinates.

Solution Summary: The author explains that the sum of the square of distances from a point P = (c, d ) to n fixed points is minimized, when c is the average of x -coordinate

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We need to show that the sum of the square of the distances from a point $P = (c, d)$ to $n$ fixed points $(a_{1}, b_{1}), ... (a_{n}, b_{n})$ is minimized, when $c$ is the average of the $x$ -coordinates $a_{i}$ and $d$ is the average of the $y$ -coordinates.

Chapter 15 Solutions

We need to show that the sum of the square of the distances from a point P = ( c , d ) to n fixed points ( a 1 , b 1 ) , ... ( a n , b n ) is minimized, when c is the average of the x -coordinates a i and d is the average of the y -coordinates.

Solution Summary: The author explains that the sum of the square of distances from a point P = (c, d ) to n fixed points is minimized, when c is the average of x -coordinate

Concept explainers

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We need to show that the sum of the square of the distances from a point P=(c,d) to n fixed points (a1,b1),...(an,bn) is minimized, when c is the average of the x-coordinates ai and d is the average of the y-coordinates.

Chapter 15 Solutions

We need to show that the sum of the square of the distances from a point $P = (c, d)$ to $n$ fixed points $(a_{1}, b_{1}), ... (a_{n}, b_{n})$ is minimized, when $c$ is the average of the $x$ -coordinates $a_{i}$ and $d$ is the average of the $y$ -coordinates.