## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises - Page 338: 23

#### Answer

$(-\frac{1}{3},\frac{4\sqrt{2}}{3})~~$ and $~~(-\frac{1}{3},-\frac{4\sqrt{2}}{3})~~$ are the points on the ellipse that are farthest from the point $(1,0)$

#### Work Step by Step

$4x^2+y^2 = 4$ $y^2 = 4-4x^2$ $y = \pm~\sqrt{4-4x^2}$ Note that $-1 \leq x \leq 1$ and $-2 \leq y \leq 2$ We can write an expression for the distance from points on the ellipse to the point $(1,0)$: $d = \sqrt{(x-1)^2+(y-0)^2}$ $d = \sqrt{(x-1)^2+(y)^2}$ $d = \sqrt{x^2-2x+1+4-4x^2}$ $d = \sqrt{-3x^2-2x+5}$ We can find the point where $d'(x) = 0$: $d(x) = \sqrt{-3x^2-2x+5}$ $d'(x) = \frac{-6x-2}{2 \sqrt{-3x^2-2x+5}} = 0$ $-6x-2 = 0$ $x = -\frac{1}{3}$ When $0 \leq x \lt -\frac{1}{3}$, then $d'(x) \gt 0$ When $x \gt -\frac{1}{3}$, then $d'(x) \lt 0$ Thus, $x=-\frac{1}{3}$ is the point where the distance $d(x)$ is a maximum. We can find $y$: $y = \pm~\sqrt{4-4x^2}$ $y = \pm~\sqrt{4-4(-\frac{1}{3})^2}$ $y = \pm~\sqrt{\frac{32}{9}}$ $y = \pm~\frac{\sqrt{32}}{3}$ $y = \pm~\frac{4\sqrt{2}}{3}$ $(-\frac{1}{3},\frac{4\sqrt{2}}{3})~~$ and $~~(-\frac{1}{3},-\frac{4\sqrt{2}}{3})~~$ are the points on the ellipse that are farthest from the point $(1,0)$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.