I did a similar question last night
just change the numbers
http://www.jiskha.com/display.cgi?id=1417144323
The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one four times as strong as the other, are placed 11 ft apart, how far away from the stronger light source should an object be placed on the line between the two sources so as to receive the least illumination?
2 answers
I=kS/d²
where
I=illumination,
k=proportional constant >0
S=source strength >0
d=distance
One light source is S, and the other is 4S.
Let d=distance from weaker source, then
11-d =distance from stronger source.
We have illumination as a function of d
I(d) = kS/d²+k(4S)/(11-d)²
To find the minimum illumination, we equate first derivative to zero and check that it is a minimum.
I'(d)=-2kS/d³+8kS/(11-d)²
Equating to zero:
2kS/d³=8kS/(11-d)²
and solve for d:
(11-d)³=4d³
11-d=4^(1/3)d
d=11/(1+4^(1/3))=4.25 feet (approx.)
Check that it is a minimum.
I"(d)=(6kS)/d^4+(24kS)/(11-d)^4
I"(4.25)=0.02994>0 therefore minimum
where
I=illumination,
k=proportional constant >0
S=source strength >0
d=distance
One light source is S, and the other is 4S.
Let d=distance from weaker source, then
11-d =distance from stronger source.
We have illumination as a function of d
I(d) = kS/d²+k(4S)/(11-d)²
To find the minimum illumination, we equate first derivative to zero and check that it is a minimum.
I'(d)=-2kS/d³+8kS/(11-d)²
Equating to zero:
2kS/d³=8kS/(11-d)²
and solve for d:
(11-d)³=4d³
11-d=4^(1/3)d
d=11/(1+4^(1/3))=4.25 feet (approx.)
Check that it is a minimum.
I"(d)=(6kS)/d^4+(24kS)/(11-d)^4
I"(4.25)=0.02994>0 therefore minimum
0 answers