A painting in an art gallery has height h and is hung so its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle \theta subtended at his eye by the painting?) (answer may depend on h and d.)
I tried method 2 but gave up about an hour ago. I do not think there is any trick, just plug and chug.
To determine the optimal distance from the wall, we need to consider the angles involved.
Let's consider the triangle formed by the observer, the lower edge of the painting, and the upper edge of the painting. This triangle is a right triangle, with the lower edge of the painting being the base and the height of the painting being the perpendicular height.
The angle subtended at the observer's eye by the painting, θ, is the angle opposite the height h in this right triangle.
From trigonometry, we know that the tangent of an angle in a right triangle is equal to the ratio of the length of the side opposite the angle to the length of the adjacent side.
In this case, the tangent of θ is h/d. Therefore, we have:
tan(θ) = h/d
To maximize the angle θ, we need to maximize the tangent of θ.
To find the maximum value of the tangent, we want to find the value of θ for which the tangent is undefined, which occurs when the adjacent side (distance from the wall) is zero.
Therefore, the observer should stand as close to the wall as possible, in other words, at a distance of zero from the wall.
To find the position where the observer should stand to get the best view, we need to maximize the angle θ subtended at their eye by the painting. Let's break down the problem step by step:
1. Draw a diagram: Draw a vertical line to represent the wall, and label it as the reference line. Now draw a horizontal line representing the lower edge of the painting, positioned at a distance d above the observer's eye level.
2. Label the points: Assign labels to the relevant points in the diagram. Let A be the lower edge of the painting, O be the observer's eye level, and P represent the position where the observer stands.
3. Determine the variables: Identify the known quantities and the variables. Here, the height of the painting is h, and the distance of the painting's lower edge from the observer's eye is d. The variable in this case is the distance from the wall where the observer stands, which we'll denote as x.
4. Establish the angle: The angle θ subtended at the observer's eye by the painting can be expressed as the angle ∠AOB in the diagram, where B represents the top edge of the painting.
5. Find the triangle: Notice that ΔAOB is a right triangle, with AB as the hypotenuse and AO and OB as the legs.
6. Apply trigonometry: We can use trigonometry to find the angle θ in terms of the given values. The tangent of the angle θ can be defined as the ratio of the opposite side (AO) to the adjacent side (OB). Therefore:
tan(θ) = AO / OB
7. Express AO and OB in terms of the known quantities: In the right triangle ΔAOB, the length of OB is x, which is the distance from the wall where the observer stands. The length of AO is h + d, as it is the sum of the height of the painting and the distance of the painting's lower edge from the observer's eye.
8. Substitute the values: Replace AO with h + d, and OB with x in the tangent equation:
tan(θ) = (h + d) / x
9. Maximize the angle θ: To maximize the angle θ, we need to find the maximum value of the tangent function. This occurs when the denominator, x, is minimized. In other words, the observer should stand as close to the wall as possible.
10. Conclusion: The observer should stand as close to the wall as possible to obtain the best view of the painting, maximizing the angle θ subtended at their eye.
Draw a diagram, looking from the side. If the observer is at distance x, then
The angle Ø from the eye to the bottom of the picture is such that
tanØ = d/x
The angle θ subtending the picture is
tan(Ø+θ) = (d+h)/x
We want to determine x for maximum θ.
So, eliminating x, we get
d/tanØ = (d+h)/tan(Ø+θ)
Nah, that's getting messy. How about this: The distance a from the eye to the bottom of the picture is
a^2 = x^2+d^2
The distance b from the eye to the top of the picture is
b^2 = x^2 + (d+h)^2
Now, we want to find maximum θ using the law of cosines:
h^2 = b^2 + a^2 - 2abcosθ
h^2 = x^2+d^2 + x^2 + (d+h)^2 - 2√((x^2+d^2)(x^2+(d+h)^2)cosθ
cosθ = (h^2-(x^2+d^2)-(x^2+(d+h)^2))/(2√((x^2+d^2)(x^2+(d+h)^2))
= -2(x^2+dh+d^2)/√(x^4 + (d^2+(d+h)^2)x^2 + d^2(d+h)^2)
Messy, but if we let c = d+h, then
cosθ = -2(x^2+dc)/√(x^4 + (d^2+c^2)x^2 + d^2c^2)
-sinθ dθ/dx = 2x(c-d)^2(cd-x^2)/(junk)^3/2
The junk is that radical to the 3/2. We don't really have to worry about it, because it is never zero, and sinθ is not zero, so we just need
2x(c-d)^2(cd-x^2) = 0
2xh^2((d+h)d-x^2) = 0
2h^2x^3 - 2dh^3x - 2d^2h^2x = 0
2x(h^2x^2 - dh(h^2+dh)) = 0
x^2 = dh^2(h+d)/h^2
whew - still messy
A view of the solution using angles is at
http://www.nfva.info/LarsonCalc/Ch5_files/ch5_5_6w_files/image018.jpg
but it has actual numbers. You will have to plug in h and d for a general solution.