Answer to Question #285996 in Calculus for musus

Question

A rectangular billboard 5 feet in height stands in a field so that its bottom is 6 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

From the diagram above, we can have that

$CD=x ,$ which is the distance of the base.

$B_1D=1, B_2D=7,$

$\theta=\beta-\alpha$

$\theta=\tan^{-1}(\frac{7}{x})-\tan^{-1}(\frac{1}{x})$

Find the derivative of $\theta$ wrt $x$

$\frac{d \theta}{dx}=-\frac{7}{x^2+49}+\frac{1}{x^2+1}$

Set the derivative to 0

$-\frac{7}{x^2+49}+\frac{1}{x^2+1}=0\\ \text{Multiply through by }(x^2+49)(x^2+1)\\ -7(x^2+1)+(x^2+49)=0\\ -7x^2-7+x^2+49=0\\ -6x^2+42=0\\ x^2=7\\ x=\sqrt{7}$

Thus the critical point is $\sqrt{7}$ . Let check if $x=\sqrt{7}$ is a maximum point

$\theta'(\sqrt{7})=-\frac{3}{56}\sqrt{7}<0$ .

Thus, $x=\sqrt{7}$ is a maximum point.

Hence the distance of the cow from the billboard must be $\sqrt{7}ft$ so as to maximize it