Answer to Question #285959 in Calculus for aniyazli

Question

A rectangular billboard 6 feet in height stands in a field so that its bottom is 13 feet above the ground. A nearsighted cow with eye level at 4 feet above the ground stands x

x

 feet from the billboard. Express θ

θ

, the vertical angle subtended by the billboard at her eye, in terms of x

x

. Then find the distance x

x

 the cow must stand from the billboard to maximize θ

θ

.

Answer 1

$$\theta(x)=\tan^{-1}\dfrac{13+6-4}{x}-\tan^{-1}\dfrac{13-4}{x}$$

$$\theta(x)=\tan^{-1}\dfrac{15}{x}-\tan^{-1}\dfrac{9}{x}, x>0$$

Take the derivative with respect to $x$

$$\theta '(x)=\dfrac{1}{1+(\dfrac{15}{x})^2}(-\dfrac{15}{x^2})-\dfrac{1}{1+(\dfrac{9}{x})^2}(-\dfrac{9}{x^2})$$

$$=\dfrac{9}{x^2+81}-\dfrac{15}{x^2+225}$$

Find the critical number(s)

$$\theta '(x)=0=>\dfrac{9}{x^2+81}-\dfrac{15}{x^2+225}=0$$

$$\dfrac{9}{x^2+81}=\dfrac{15}{x^2+225}$$

$$3(x^2+225)=5(x^2+81)$$

$$2x^2=270$$

$$x^2=135$$

$$x=\pm\sqrt{135}$$

$$x=\pm3\sqrt{15}$$

Since $x>0$ we take $x=3\sqrt{15}.$

If $0<x<3\sqrt{15}, \theta'(x)>0, \theta(x)$ increases.

If $x>3\sqrt{15}, \theta'(x)<0, \theta(x)$ decreases.

The cow must stand at $3\sqrt{15}$ feet from the billboard to maximize $θ.$