Answered: consider the function f(x) = 12(x-1)^-2…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

consider the function f(x) = 12(x-1)^-2

find the value c such that the triangle formed by the tangent line to f at c and the axes has the largest possible area

c=______________________

find the maximum area of the triangle formed by the tangent line to f at c and the axes

A=_______________________________units^2

Expert Solution

Step 1

Equation of tangent is:

$\frac{y - f (c)}{x - c} = {\frac{d}{d x} f (x)}_{x = c} \frac{y - \frac{12}{{(c - 1)}^{2}}}{x - c} = \frac{- 2 \times 12}{{(c - 1)}^{3}} \frac{y {(c - 1)}^{2} - 12}{x - c} = \frac{- 24}{c - 1}$

Step 2

Point wherein the tangent will touch x-axis is y=0 or:

$\frac{0 . {(c - 1)}^{2} - 12}{x - c} = \frac{- 24}{c - 1} x = \frac{(c - 1)}{2} + c x = \frac{3 c - 1}{2}$

Step 3

Point wherein the tangent will touch y-axis is x=0 or:

$\frac{y . {(c - 1)}^{2} - 12}{0 - c} = \frac{- 24}{c - 1} y . {(c - 1)}^{2} - 12 = \frac{24 c}{c - 1} y . {(c - 1)}^{2} - 12 = \frac{24 (c - 1 + 1)}{c - 1} y . {(c - 1)}^{2} - 12 = 24 + \frac{24}{c - 1} y . {(c - 1)}^{2} = 36 + \frac{24}{c - 1} y = \frac{36}{{(c - 1)}^{2}} + \frac{24}{{(c - 1)}^{3}} y = \frac{36 (c - 1) + 24}{{(c - 1)}^{3}} y = \frac{12 (3 c - 1)}{{(c - 1)}^{3}}$

Step 4

Thus area of triangle is $A = \frac{1}{2} x y$ or $A (c) = \frac{1}{2} (\frac{12 (3 c - 1)}{{(c - 1)}^{3}}) (\frac{3 c - 1}{2})$ or $A (c) = \frac{3 {(3 c - 1)}^{2}}{{(c - 1)}^{3}}$ .

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