Question

the illumination due to a source of light varies directly as the strength of the source and inversely as the square of the distance from the source. two sources are 6 meters apart and one of them is 8 times stronger than the other. at what distance from the weaker source on the line segment joining them is the illumination the least

Answer 1

Point is at x = 4 m

Explanation:

From the question, we can write the formula for intensity as;

I = strength/distance²

Now,

- Let x be the position from the stronger source

- Let k be the strength of the weaker light source

- 8k will be the strength of the stronger light source.

We are told that the two sources are 6 meters apart and one of them is 8 times stronger than the other.

Thus, the total illumination is;

I(x) = (8k/x²) + k/(6 - x)²

Using chain rule to get the critical points, let's find the first derivative and equate to zero;

I'(x) = -16k/x³ + 2k/(6 - x)³ = 0

Adding -16k/x³ to both sides, we have;

2k/(6 - x)³ = 16k/x³

Cross multiply to get;

2kx³ = 16k(6 - x)³

Dividing both sides by 16k to give;

x³/8 = (6 - x)³

Taking cube root of both sides to give;

x/2 = 6 - x

Multiply both sides by 2 to give;

x = 12 - 2x

x + 2x = 12

3x = 12

x = 12/3

x = 4 m