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 $8\text{ft}$ tell runs parallel to a tall building at a distance of $4\text{ft}$ from the building.

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 $16.65\text{ft}$

Explanation of Solution

Given information:

A fence $8\text{ft}$ tell runs parallel to a tall building at a distance of $4\text{ft}$ 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:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

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

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

  

  $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AB}{QR}$

And

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

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • 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 $8\text{ft}$ tell runs parallel to a tall building at a distance of $4\text{ft}$ from the building

  

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

Corresponding sides of the triangles are in proportional,

Here, $△ADE\text{and }△ABC$ are similar,

Therefore,

  $\begin{array}{l}\frac{h}{8}=\frac{x+4}{x}\\ \Rightarrow h=\frac{8\left(x+4\right)}{x}\end{array}$

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 $△ABC$

  $l^2=h^2+{\left(x+4\right)}^2$

Substitute $h=\frac{8\left(x+4\right)}{x}$ ,and simplified

  $\begin{array}{l}{l}^2=h^2+{\left(x+4\right)}^2\\ l^2\left(x\right)={\left\{\frac{8\left(x+4\right)}{x}\right\}}^2+{\left(x+4\right)}^2\\ l^2\left(x\right)={\left\{8+\frac{32}{x}\right\}}^2+{\left(x+4\right)}^2\mathrm{......}\left(1\right)\end{array}$

Recall that,

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

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ 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.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • 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,

  $\begin{array}{l}\frac{d}{dx}{l}^2\left(x\right)=\frac{d}{dx}\left[{\left(8+\frac{32}{x}\right)}^2+{\left(x+4\right)}^2\right]\\ =2\left(8+\frac{32}{x}\right)\times -\frac{32}{x^2}+2\left(x+4\right)\times 1\\ =-64\left(\frac{32+8x}{x^3}\right)+2\left(x+4\right)\mathrm{......}\left(2\right)\end{array}$

Solve for $\frac{d}{dx}{l}^2\left(x\right)=0$ , and simplified

  $\begin{array}{l}-64\left(\frac{32+8x}{x^3}\right)+2\left(x+4\right)=0\\ \Rightarrow 64\times 8\left(\frac{4+x}{x^3}\right)=2\left(x+4\right)\\ \Rightarrow \frac{1}{x^3}=\frac{1}{256}\\ \Rightarrow x^3=256\end{array}$

Take cube root on both sides,

  $\begin{array}{l}x=\sqrt[3]{256}\\ \Rightarrow x\approx 6.35\end{array}$

Differentiate equation $\left(2\right)$ with respect to $l$

  $\begin{array}{l}\frac{d^2}{d{x}^2}{l}^2\left(x\right)=-64\frac{d}{dx}\left(\frac{32+8x}{x^3}\right)+2\frac{d}{dx}\left(x+4\right)\\ =-64\left(-\frac{96}{x^4}-\frac{16}{x^2}\right)+2\\ =64\left(\frac{96}{x^4}+\frac{16}{x^2}\right)+2\end{array}$

Therefore,

  $\frac{d^2}{d{x}^2}{l}^2\left(x\right)=64\left(\frac{96}{x^4}+\frac{16}{x^2}\right)+2\ge 0$

For $x\approx 6.35$ the length of ladder is minimum

Substitute $x\approx 6.35$ in equation $\left(1\right)$ and simplified

  $\begin{array}{l}{l}^2\left(6.35\right)={\left\{8+\frac{32}{6.35}\right\}}^2+{\left(6.35+4\right)}^2\\ \approx 169.78+107.12\\ \approx 276.90\end{array}$

Take square root on both sides,

  $l\left(6.35\right)\approx 16.65$

Conclusion:

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