Question

Carrying a Ladder Around a Corner Two hallways, one of width 3 feet, the other of width 4 feet, meet at a right angle. See the illustration. It can be shown that the length L of the ladder as a function of $\theta$is $L\left(\theta \right)=4csc\theta +3sec\theta .$

(a) In calculus, you are asked to find the length of the longest ladder that can turn the corner by solving the equation

$3sec\theta \tan \theta -4csc\theta cot\theta =0,0°<\theta <90°$

Solve this equation for $\theta .$

(b) What is the length of the longest ladder that can be carried around the corner?

(c) Graph $L=L\left(\theta \right),0°\le \theta \le 90°,$and find the angle $\theta$ that minimizes the length L.

(d) Compare the result with the one found in part (b). Explain why the two answers are the same.

Short answer

Part a. The solution of the equation in $\theta \mathrm{is}47.7°.$

Part b. The length of the longest ladder that can be carried around the corner is $9.86\mathrm{feet}.$

Part c. The graph of$L=L\left(\theta \right),0°\le \theta \le 90°$is

The angle is $\theta =47.7°$that minimize the length L.

Part d. The two answers are the same because there is no maximum point on the graph, thus the minimum point will be assumed as the length.

Step-by-step solution

Part (a) Step 1. Given Information

There are two hallways of width $3\mathrm{and}4$feet, meet at a right angle.

The length of the ladder as a function is $L\left(\theta \right)=4csc\theta +3sec\theta .$

We have to solve the equation$3sec\theta \tan \theta -4csc\theta cot\theta =0,0°<\theta <90°.$

Part (a) Step 2. Simplifying the equation

We have to find the length of the longest ladder that can turn the corner by solving the equation $3sec\theta \tan \theta -4csc\theta cot\theta =0,0°<\theta <90°$.

Thus,

role="math" localid="1647349344317" $$\begin{gathered} 3sec\theta \tan \theta -4csc\theta cot\theta =0 \\ 3\left(\frac{1}{\cos \theta }\right)\left(\frac{\sin \theta }{\cos \theta }\right)-4\left(\frac{1}{\sin \theta }\right)\left(\frac{\cos \theta }{\sin \theta }\right)=0 \\ \frac{3\sin \theta }{{\cos }^2\theta }-\frac{4\cos \theta }{{\sin }^2\theta }=0 \\ \frac{3\sin \theta }{{\cos }^2\theta }=\frac{4\cos \theta }{{\sin }^2\theta } \\ \frac{{\sin }^3\theta }{{\cos }^3\theta }=\frac{4}{3} \\ {\tan }^3\theta =1.333 \\ \tan \theta ={\left(1.333\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \tan \theta =1.1 \\ \theta ={\tan }^{-1}\left(1.1\right) \\ \theta =47.7° \end{gathered}$$

Part (b) Step 1. Finding the length of the longest ladder

To find the length of the longest ladder that can be carried around the corner we will use the function of the ladder $L\left(\theta \right)=4csc\theta +3sec\theta$

Now, the length of the ladder is longest when $\theta =47.7°$

$$\begin{gathered} L_{Max}\left(\theta \right)=4csc\left(47.7°\right)+3sec\left(47.7°\right) \\ {L}_{Max}\left(\theta \right)=4\left(1.352\right)+3\left(1.485\right) \\ {L}_{Max}\left(\theta \right)=5.408+4.46 \\ {L}_{Max}\left(\theta \right)=9.86 \end{gathered}$$

So, the length of the longest ladder is$9.86\mathrm{feet}.$

Part (c) Step 1. Sketch the graph

The graph of $L=L\left(\theta \right),0°\le \theta \le 90°$is

The angle that minimizes the length is$47.7°.$

Part (d) Step 4. Comparing the result

From part (b) we get the length of the longest ladder is $9.86\mathrm{feet}$and from part (c) we get that the angle that minimizes the length is at the same feet we get from part (b). That's why the answer is the same and there is no maximum point on the graph, thus the minimum point will be assumed as the length.