## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1009: 59

#### Answer

$m\Sigma_{i=1}^{n}x_i+bn=\Sigma_{i=1}^{n} y_i$ and $m\Sigma_{i=1}^{n}x_i^2+b\Sigma_{i=1}^{n}x_i=\Sigma_{i=1}^{n}x_i y_i$

#### Work Step by Step

Given: $d_i=y_i-(mx_i+b)$ These are the vertical deviation of the point $(x_i,y_i)$ from the straight line. Now, $\dfrac{\partial L}{\partial m}=\Sigma_{i=1}^{n}\dfrac{\partial }{\partial m}[y_i-(mx_i+b)]$ or, $=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-x_i)$ or, $\dfrac{\partial L}{\partial m}=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-x_i)$ Thus, $m\Sigma_{i=1}^{n}x_i^2+b\Sigma_{i=1}^{n}x_i=\Sigma_{i=1}^{n}x_i y_i$ Also, Given: $d_i=y_i-(mx_i+b)$ $\dfrac{\partial L}{\partial b}=\Sigma_{i=1}^{n}\dfrac{\partial }{\partial m}[y_i-(mx_i+b)]$ or, $=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-1)$ or, $\dfrac{\partial L}{\partial b}=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-1)$ Thus, $m\Sigma_{i=1}^{n}x_i+bn=\Sigma_{i=1}^{n} y_i$

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