Answer to Question #280856 in Calculus for kenneth

Question

A wall "h" meters high is 2m away from the building. The shortest ladder that can reach the building with one end resting on the ground outside the wall is 6m. How high is the wall in meters?

Answer 1

Let $L=x+y$ be the lengh of the ladder.

$$\cos \theta=\dfrac{2}{y}, \sin \theta=\dfrac{h}{x}$$

Then

$$L=x+y=\dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta}$$

$$L=\dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta}$$

Find the first derivative with respect to $\theta$

$$\dfrac{dL}{d\theta}=-\dfrac{h\cos \theta}{\sin^2\theta}+\dfrac{2\sin \theta}{\cos^2 \theta}$$

Find the critical number(s)

$$\dfrac{dL}{d\theta}=0=>-\dfrac{h\cos \theta}{\sin^2\theta}+\dfrac{2\sin \theta}{\cos^2 \theta}=0$$

$$h\cos^3\theta=2\sin^3\theta$$

$$h=2\tan^3\theta$$

Given that  the shortest ladder is 6 m.

Substitute

$$\dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta}=6$$

$$2\tan^3\theta\cos \theta+2\sin \theta=6\sin \theta\cos \theta$$

$0\degree<\theta< 90\degree$

$$\sin^3\theta+\sin \theta\cos^2\theta=3\sin \theta\cos^3\theta$$

$$1=3\cos^3\theta$$

$$\cos^3\theta=\dfrac{1}{3}$$

$$\cos \theta=\dfrac{1}{\sqrt[3]{3}}$$

$$\sin \theta=\sqrt{1-\dfrac{1}{\sqrt[3]{3^2}}}=\dfrac{\sqrt{\sqrt[3]{9}-1}}{\sqrt[3]{3}}$$

$$h=2\bigg(\sqrt{\sqrt[3]{9}-1}\bigg)^3$$

$$h\approx2.245\ m$$