333
Answers
501–600
Answers and Explanations
Let x be the distance of the object from the brighter light source and k be the strength
of the weaker light source. The illumination of each light source is directly proportional to the strength of the light source (5k and k) and inversely proportional to the
square of the distance from the source (x and 20 – x), so the total illumination is
I x
k
x
k
x
( )
(
)
=
+
−
5
20
2
2
where 0 < x < 20.
Taking the derivative of this function gives you
′
I x
k
x
k
x
( )
(
)
= −
+
−
10
2
20
3
3
Set this derivative equal to zero and simplify:
−
−
−
+
−
=
=
−
=
10 20
20
2
20
0
2
10 20
5 20
3
3
3
3
3
3
3
3
3
k
x
x
x
kx
x
x
kx
k
x
x
(
)
(
)
(
)
(
)
( − − x )
3
Then solve for x. Taking the cube root of both sides and solving gives you
x
x
x
x
x
x
x
x
x
x
3
3
3
3
3
3
3
3
3
5 20
5 20
20 5
5
5
20 5
1 5
20 5
20 5
=
−
=
−
=
−
=
=
=
(
)
(
)
+
+
(
)
3 3
3
1 5
12 62
+
≈ . ft
Note that you can use the first derivative test to verify that this value does in fact give
a minimum.
512.
12