# A picture frame 5 ft. high hangs on a vertical wall so that the bottom edge is 4 ft. above an observer’s eye. How far (in ft.) should the observer be from the wall so that the vertical angle at the eye subtended by the picture, is greatest.

## If the eye is x ft from the wall, then we have

θ = angle subtended by frame
Ø = angle to bottom of frame
Now consider
tanØ = 4/x
9/x = tan(θ+Ø) = (tanθ + tanØ)/(1 - tanθ tanØ)
9/x = (tanθ + 4/x)/(1 - tanθ * 4/x)
9/x - 36/x^2 tanθ = tanθ + 4/x
tanθ = 5x/(x^2+36)
sec^2θ dθ/dx = -5(x^2-36)/(x^2+36)^2
dθ/dx = 1/(tan^2θ + 1) * -5(x^2-36)/(x^2+36)^2
= -5(x^2-36)/(((5/(x^2+36))^2+1)(x^2+36)^2)
= -5(x^2-36)/(x^4+72x^2+1321)
since the denominator is never zero, dθ/dx=0 when x=6

## To find the distance from the wall at which the vertical angle at the eye subtended by the picture is greatest, we can use trigonometry.

Let's label the distance from the observer to the wall as "x" (in ft).

Since the bottom edge of the frame is 4 ft above the observer's eye, the height of the frame from the observer's eye level is 5 ft - 4 ft = 1 ft.

Using trigonometry, we can determine the angle "θ" subtended by the picture at the observer's eye.

tan(θ) = 1 ft / x
θ = tan^(-1)(1 ft / x)

To find the value of x that maximizes the angle θ, we can find the maximum value of the tangent function.

The tangent function reaches its maximum value at π/2 radians or 90 degrees.

So, θ will be maximized when tan^(-1)(1 ft / x) is equal to π/2, or 90 degrees.

tan^(-1)(1 ft / x) = π/2

Using trigonometric identities, we know that tan^(-1)(1) = π/4

So, we have:

π/4 = π/2

1 ft / x = 1

x = 1 ft

Therefore, the observer should be 1 ft away from the wall to maximize the vertical angle at the eye subtended by the picture.

## To find the distance at which the vertical angle at the eye subtended by the picture is greatest, we need to use trigonometry.

Let's denote the distance from the observer to the wall as "x". We can set up a right triangle where the height of the picture frame (5 ft) is the opposite side of the right angle, and the distance from the bottom edge of the frame to the observer's eye (4 ft) is the adjacent side of the right angle.

Using the tangent function, we can calculate the vertical angle at the eye, which is equal to the angle between the observer's line of sight and the top edge of the frame:

tan(θ) = 5 / x

Now we want to find the value of "x" that maximizes the vertical angle θ. In other words, we are looking for the maximum value of tan(θ).

To find the maximum value of tan(θ), we can take the derivative of the equation with respect to "x" and set it equal to zero. Then solve the resulting equation for "x".

d/dx (tan(θ)) = 0
d/dx (5 / x) = 0
-5 / x^2 = 0

Since the derivative equals zero when the denominator is zero, we conclude that x = 0. However, this value is not meaningful in this context.

Next, we consider the endpoints of the domain. As the observer moves away from the wall, "x" gets larger. As x approaches infinity, the vertical angle θ approaches a maximum value, which is 90 degrees.

Therefore, the vertical angle at the eye subtended by the picture is greatest when the observer is at an infinite distance from the wall.