SOLUTION: Can anyone show me how to solve this? Thanks in advance. Tony buys two sun lamps. The intensity of a sun lamp is inversely proportional to the square of the distance from the la

Question 22000: Can anyone show me how to solve this? Thanks in advance.
Tony buys two sun lamps. The intensity of a sun lamp is inversely proportional to the square of the distance from the lamp. When he gets home, he discovers he has purchased one 50 watt lamp and one 60 watt lamp. He positions the lamps exactly 10 meters apart. He then sits, between the lamps, on an imaginary line running directly between the two lamps. To ensure he gets an even tan, how far d should he sit from the 50 watt lamp?
d=______m
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
LET A BE THE POSITION OF 50 WATT LAMP AND B BE THE POITION OF THE 60 WATT LAMP
AB=10 M.
LET TONY SIT AT T BETWEEN A AND B.LET AT=X METRES.SO TB=10-X METRES.
INTENSITY IS INVERSELY PROPORTIONAL TO THE DISTANCE FROM THE LAMP.IF WE DENOTE INTENSITY BY I AND DISTACS BY D THEN THIS MEANS
I = K*W/X^2...WHERE K IS A CONSTANT AND W IS THE WATTAGE OF THE LIGHT.(NOTE THAT IF P IS DIRECTLY PROPORTIONAL TO Q THEN P=K*Q,WHERE K IS CONSTANT AND IF P IS INVERSELY PROPORTIONAL TO Q THEN P=K/Q)
NOW THE INTENSITY AT T WHERE TONY SITS IS EQUAL FROM BOTH THE LIGHTS
INTENSITY FROM 50 WATT LAMP =K*50/X^2
INTENSITY FROM 60 WATT LAMP =K*60/(10-X)^2...THESE ARE EQUAL.HENCE
K*50/X^2=K*60/(10-X)^2....CROSS MULTIPLYING..
50*(10-X)^2=60*X^2
50=60*X^2
60*X^2-50*X^2+1000X-5000=0
10X^2+1000X-5000=0
X^2+100X-500=0
USING THE FORMULA FOR ROOTS OF A QUADRATIC EQN.ax^2+bx+c=0

x =4.77

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