Question

You are the coordinator for a program that is going to take place at night in a rectangular amphitheater in the mountains. You will have no access to any electricity, but you must be able to illuminate the entire grounds. You know the intensity of the light from a lantern varies inversely as the square of the distance from the lantern. Suppose the intensity is 90 when the distance is 5 m.
a. Write an equation to model the situation.
b. Solve for the constant of variation.
c. Write the equation to model the situation using the constant () of variation.
d. You have been given lanterns with 40 light intensity. Use your equation to solve for the distance from the lantern.
e. You need to illuminate 225 km. How many meters do you need to light?
f. How many lanterns will you need?

Answer 1

The equation to model the situation is $\mathbf{y = \dfrac{k}{x^2}}$. The constant for the variation is 2250.

What is the intensity of light?

The intensity of light from a lantern varies inversely to the square of the distance from the lantern.

From the given information:

• Let y be the intensity of light, and
• x be the distance from the lantern

Then:

$\mathbf{y \alpha \dfrac{1}{x^2} }$

$\mathbf{y = \dfrac{k}{x^2} }$   here, k = constant.

2.

If y = 90 W/m² when the distance x = 5m

Then:

$\mathbf{90 = \dfrac{k}{(5)^2}}$

k = 90 × 25

k = 2250

c.

The equation to model the situation by using the constant variation is:

$\mathbf{y = \dfrac{2250}{x^2}}$

d.

If the light intensity y = 40, then x is determined as:

$\mathbf{40 = \dfrac{2250}{x^2}}$

$\mathbf{x = \sqrt{\dfrac{2250}{40}}}$

x = 7.5 m

e.

The light is needed in (225 × 1000)m = 225000 km of illumination.

f.

The lantern required for the new light estimation is:

y = 2250/225000

y = 0.01 intensity

Therefore, we can conclude that to get an intensity of 1 W/m², we need to put 100 lanterns.

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