A light is suspended at a height h above the floor. The illumination at the point P is inversely proportional to the square of the distance from the point P to the light and directly proportional to the cosine of the angle θ. How far from the floor should the light be to maximize the illumination at the point P? (Let d = 6 ft.)
I guess d is from below the light to a point on the floor?
If so:
I = k cos T / (h^2+d^2)
where cos T = h/(h^2+d^2)^.5
so
I = k h/(h^2+d^2)^1.5
dI/dh = k [ (h^2+d^2)^1.5 -h(1.5)(h^2+d^2)^.5 (2h) ] / (h^2+d^2)^3
= 0 for extreme
(h^2+d^2)^1.5 = 3 h^2 (h^2+d^2)^.5
(h^2+d^2) = 3 h^2
d^2 = 2 h^2
d = h sqrt 2
h = 6/sqrt 2 = 4.24 ft
To find the distance from the floor at which the light should be placed to maximize the illumination at point P, we need to determine an expression for the illumination at P in terms of the distance from the floor.
Let's start by visualizing the problem. We have a light source suspended at a height h above the floor, and we want to find the optimal height that maximizes the illumination at point P, which is at a distance d from the floor.
The illumination at point P is given by two factors: inversely proportional to the square of the distance from P to the light source (let's call it x), and directly proportional to the cosine of the angle θ.
First, let's express the illumination I as a function of x and θ:
I = k * (1/x^2) * cos(θ)
where k is a constant of proportionality.
Since we are interested in maximizing the illumination at point P, we can treat I as the dependent variable and x as the independent variable. Let's express I solely as a function of x:
I(x) = k/x^2 * cos(θ)
Now, we need to find a relation between x, h, and d. From the problem statement, we know that x^2 = h^2 + d^2. Therefore, we can express cos(θ) as follows:
cos(θ) = d / sqrt(h^2 + d^2)
Substituting this into the expression for I(x):
I(x) = k/x^2 * (d / sqrt(h^2 + d^2))
To maximize I(x), we should find the critical points of this function. Taking the derivative with respect to x and setting it equal to zero, we can find the maximum:
dI/dx = 0
To solve this equation, we'll differentiate the expression for I(x) with respect to x:
dI/dx = -2k/x^3 * (d / sqrt(h^2 + d^2)) + k/x^2 * ((-1/2) * (h^2 + d^2)^(-3/2)) * 2xd
Set this derivative equal to zero and solve for x:
-2k/x^3 * (d / sqrt(h^2 + d^2)) + k/x^2 * ((-1/2) * (h^2 + d^2)^(-3/2)) * 2xd = 0
Simplifying, we get:
-2k * (d / sqrt(h^2 + d^2)) + k * ((-1/2) * (h^2 + d^2)^(-3/2)) * 2xd = 0
Now, we can solve for x:
-2k * (d / sqrt(h^2 + d^2)) = k * ((-1/2) * (h^2 + d^2)^(-3/2)) * 2xd
Simplifying further:
-2d = -xd / sqrt(h^2 + d^2)
Now we can solve for x:
2 / sqrt(h^2 + d^2) = x / d
Cross-multiplying:
2d = xd
2 = x
Therefore, to maximize the illumination at point P, the distance from the floor (x) should be equal to 2.
Thus, the light should be placed 2 feet above the floor to maximize the illumination at point P.
To find the height at which the light should be suspended to maximize the illumination at point P, we can use calculus.
Let's denote the height at which the light is suspended as x. We want to maximize the illumination at point P, which is inversely proportional to the square of the distance from P to the light and directly proportional to the cosine of the angle θ.
1. Begin by finding the distance r from the light to point P.
r = √(h^2 + x^2)
2. Next, find the angle θ between the horizontal line and the line connecting point P and the light.
cos θ = x / r
cos θ = x / √(h^2 + x^2)
3. The illumination I at point P is then given by:
I = k / (r^2 * cos θ)
I = k / [(h^2 + x^2) * (x / √(h^2 + x^2))^2]
I = k / [x^2 * (√(h^2 + x^2))^2]
I = k / (x^2 * (h^2 + x^2))
Note: k is a constant of proportionality that we don't need to worry about when maximizing the illumination.
4. To maximize I, we need to find the critical points by taking the derivative of I with respect to x and setting it equal to zero.
dI/dx = 0
By simplifying and setting the numerator equal to zero:
2x(h^2 + x^2) - 2x^3 = 0
x(h^2 + x^2) - x^3 = 0
x(h^2 - x^2) = 0
x = 0 or x = √h
We discard x = 0 since it does not make sense physically in this context.
5. We need to determine if this critical point is a maximum or minimum. We can do this by taking the second derivative of I with respect to x.
d^2I/dx^2 = 2(h^2 + x^2) - 6x^2
6. Plug in x = √h into the second derivative:
d^2I/dx^2 at x = √h = 2(h^2 + (√h)^2) - 6(√h)^2
= 2(h^2 + h) - 6h
= 2h^2 + 2h - 6h
= 2h^2 - 4h
7. Evaluate the second derivative at x = √h:
d^2I/dx^2 = 2h^2 - 4h
8. Determine if this value is positive or negative to determine if x = √h is a maximum or minimum. Since h is positive,
d^2I/dx^2 = 2h^2 - 4h > 0 (for h > 0)
Therefore, our critical point x = √h is a local minimum.
Since we only have one critical point and it is a minimum, this means it is our global maximum.
Therefore, the light should be suspended at a height of √h from the floor to maximize the illumination at point P.