Q. 51 - Precalculus Enhanced with Graphing Utilities

Question

Two hallways, one of width $3$ feet, the other of width $4$ feet, meet at a right angle.

See the illustration.

(a) Show that the length L of the line segment shown as a function of the angle $\theta$is

$L\left(\theta \right)=3 sec \theta +4 csc \theta$

(b) Graph $L=L\left(\theta \right),0<\theta <\frac{\mathrm{\pi }}{2}$

(c) For what value of $\theta$ is L the least?

(d) What is the length of the longest ladder that can be carried

around the corner? Why is this also the least value of L?

Step-by-Step Solution

Answer

According to the diagram

(a) Function for the length of the ladder is $L\left(\theta \right)=3 sec \theta +4 csc \theta$

(b) Graph of $L=L\left(\theta \right),0<\theta <\frac{\mathrm{\pi }}{2}$is

(c) The least value of L is at $\theta =0.83$

(d) the length of the longest ladder is$9.866 ft$

Step 1. Given data

Diagram of two hallway 

Function for Ladder length is

$L\left(\theta \right)=3 sec \theta +4 csc \theta$

Step 2. Diagram

Plot the diagram and label it

Step 3. Part (a)

In triangle $∆oab$

$$\begin{gathered} csc (90-\theta )=\frac{ao}{ab} \\ csc (90-\theta )=\frac{ao}{3} \\ sec \left(\theta \right)=\frac{ao}{3} \\ 3sec \left(\theta \right)=ao \end{gathered}$$

In triangle$∆omn$

$$\begin{gathered} csc \theta =\frac{no}{nm} \\ csc \theta =\frac{no}{4} \\ 4csc \theta =no \end{gathered}$$

Length of ladder

$$\begin{gathered} L=ao+no \\ L\left(\theta \right)=3 sec \theta +4 csc \theta \end{gathered}$$

Step 4. Part (b)

Plot the graph of $L\left(\theta \right)=3 sec \theta +4 csc \theta$

Step 5. Part (c)

The graph of $L\left(\theta \right)=3 sec \theta +4 csc \theta$ shows that the minima of the function is at $(0.83,9.88)$

So, L is least at $\theta =0.83 radian$

Step 6. Part (d)

As the angle reaches to $0$ or $\frac{\mathrm{\pi }}{2}$, the length of the ladder reaches infinity

so therefore the length of the longest ladder that can be carried around the corner corresponds to the least value of $\theta$

As the known ladder is least at $\theta =0.83$

Substitute the $0.83$ for $\theta$ in the function

$$\begin{gathered} L(0.83)=3 sec (0.83)+4 csc (0.83) \\ \approx 9.866 \end{gathered}$$

So longest ladder is $9.866 ft$