Answer to Question #275063 in Calculus for Hetal

Question

n the right triangle ABC, AB = 2, BC = 4 and ED is a line parallel to AB. Find the

angle α = angle BAD which minimizes the distance L, where L = AD + ED

Answer 1

Let $BD=x.$

Right triangle $ABD$

$BD=AB\tan \alpha, AD=\dfrac{AB}{\cos \alpha}$

Given $AB = 2, BC = 4.$

Substitute

$x=2\tan \alpha, AD=\dfrac{2}{\cos \alpha}$

Since $ED\parallel AB,$ then right triangles $ABC$ and $ABD$ are similar.

$\dfrac{ED}{AB}=\dfrac{CD}{BC}=>ED=\dfrac{AB}{BC}\cdot CD$

$CD=BC-BD=4-2\tan \alpha$

Substitute

$ED=\dfrac{2}{4}(4-2\tan \alpha)=2-\tan \alpha$

Then

$L=AD+ED=\dfrac{2}{\cos \alpha}+2-\tan \alpha,$ $0\degree<\alpha<90\degree$

Find the first derivative with respect to $\alpha$

$L'_{\alpha}=-\dfrac{2}{\cos ^2 \alpha}(-\sin \alpha)-\dfrac{1}{\cos ^2 \alpha}$

$=\dfrac{2\sin \alpha-1}{\cos ^2 \alpha}$

Find the critical number(s)

$L'_{\alpha}=0=>\dfrac{2\sin \alpha-1}{\cos ^2 \alpha}=0$

$\sin \alpha=\dfrac{1}{2}, 0\degree<\alpha<90\degree$

The critical number is $30\degree.$

If $0\degree<\alpha<30\degree, L'_{\alpha}<0, L$ decreases.

If $30\degree<\alpha<90\degree, L'_{\alpha}>0, L$ increases.

The function $L$ has a local minimum at $\alpha=30\degree.$

Since the function $L$ has the only extremum for $0\degree<\alpha<90\degree,$ then the function $L$ has the absolute minimum for $0\degree<\alpha<90\degree$ at $\alpha=30\degree.$

$\angle BAD=\alpha=30\degree.$