Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.6, Problem 26E

To determine

To calculate: The length of the shortest ladder that will reach from the ground over the fence to the well of the building. If fence $8 ft$ tell runs parallel to a tall building at a distance of $4 ft$ from the building.

Expert Solution & Answer

Answer to Problem 26E

The length of the shortest ladder that will reach from the ground over the fence to the well of the building $16.65 ft$

Explanation of Solution

Given information:

A fence $8 ft$ tell runs parallel to a tall building at a distance of $4 ft$ from the building.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 26E , additional homework tip 1

$(H)2=(P)2+(B)2$

Similarity property: If two triangles say $△ABC and △PQR$ are similar. Then

Corresponding sides of the triangles are in proportional, i.e

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 26E , additional homework tip 2

$ABPQ=ACPR=ABQR$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f′(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
- If $f″(a)>0$ then $f$ has a local minimum at $x=a$
- If $f″(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of a fence $8 ft$ tell runs parallel to a tall building at a distance of $4 ft$ from the building

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 26E , additional homework tip 3

Recall that, If two triangles say $△ABC and △PQR$ are similar. Then

Corresponding sides of the triangles are in proportional,

Here, $△ADE and △ABC$ are similar,

Therefore,

$h8=x+4x⇒h=8(x+4)x$

Recall that, Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle.

In $△ABC$

$l2=h2+(x+4)2$

Substitute $h=8(x+4)x$ ,and simplified

$l2=h2+(x+4)2l2(x) ={8(x+4)x}2+(x+4)2l2(x) ={8+32x}2+(x+4)2 ......(1)$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
- If $f″(a)>0$ then $f$ has a local minimum at $x=a$
- If $f″(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

$ddxl2(x) =ddx[(8+32x)2+(x+4)2] =2(8+32x)×−32x2+2(x+4)×1 =−64(32+8xx3)+2(x+4) ......(2)$

Solve for $ddxl2(x) =0$ , and simplified

$−64(32+8xx3)+2(x+4)=0⇒64×8(4+xx3)=2(x+4)⇒ 1x3=1256⇒ x3=256$

Take cube root on both sides,

$x=2563⇒ x≈6.35$

Differentiate equation $(2)$ with respect to $l$

$d2dx2l2(x) =−64ddx(32+8xx3)+2ddx(x+4) =−64(−96x4−16x2)+2 =64(96x4+16x2)+2$

Therefore,

$d2dx2l2(x) =64(96x4+16x2)+2≥0$

For $x≈6.35$ the length of ladder is minimum

Substitute $x≈6.35$ in equation $(1)$ and simplified

$l2(6.35) ={8+326.35}2+(6.35+4)2 ≈169.78+107.12 ≈276.90$

Take square root on both sides,

$l(6.35) ≈16.65$

Conclusion:

Thus the length of the shortest ladder that will reach from the ground over the fence to the well of the building $16.65 ft$

Chapter 4.6, Problem 25E

Chapter 4.6, Problem 27E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The length of the shortest ladder that will reach from the ground over the fence to the well of the building. If fence 8 ft tell runs parallel to a tall building at a distance of 4 ft from the building.

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Answer to Problem 26E

Explanation of Solution

Chapter 4 Solutions