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 feet from the billboard. Express θ, the vertical angle subtended by the billboard at her eye, in terms of x. Then find the distance x0 the cow must stand from the billboard to maximize θ.

Answer 1

Answer:

θ = $tan^{-1}$[7/x] - $tan^{-1}$[2/x]

x0 = 3.74 feet

Explanation:

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 feet from the billboard. Thus:

The vertical angle subtended by the billboard at cow's eye (θ) is equal to the difference between the angles subtended by the billboard at the bottom (θ$_{b}$) and top (θ$_{t}$) of the cow's eye level.

θ = θ$_{t}$ - θ$_{b}$ = $tan^{-1}$[(5+6-4)/x] - $tan^{-1}$[(6-4)/x]  =  $tan^{-1}$[7/x] - $tan^{-1}$[2/x]

To determine the distance x0 that maximizes θ. We will conduct the derivative with respect to the variable 'x'. Therefore;

dθ/dx = [1/(1+(7/x)^2)](-7/x)^2 -[1/(1+(2/x)^2)](-2/x)^2]

Rearranging the equation, we have:

7x^2 (1+4/x^2) = 2x^2 (1+49/x^2)

7(x^2 + 4) = 2(x^2 + 49)

7x^2 - 2x^2 = 98 - 28

5x^2 = 70

x^2 = 14

x0 = sqrt (14) = 3.74 feet

Answer 2

Answer:

In order to maximize $\theta$, the cow must stand 3.74 ft away from the billboard, approximately.

Explanation:

The vertical angle subtended by the billboard at the cow's eye can be calculated from

$\theta = \arctan{\frac{7}{x}} - \arctan{\frac{2}{x}}.$

Now, maximaxing $\theta$, we have

$\theta_{max} = \frac{d\theta}{dx}|_{x=x_0}=0\\-\frac{7}{x_0^2+49} - \left(\frac{2}{x_0^2+4}\right) = 0\\\frac{7}{x_0^2+49} - \frac{2}{x_0^2+4} = 0.$

Solving the equation for $x_0$

$x_0 = \pm\sqrt{14} \approx 3.74.$

Which can be interpretated as the cow standing either one or the other side from the billboard. The next plot confirms that $x_0$'s positive root, gives a maximum for the angle $\theta$.