The electrical power, P, in watts produced by a light is given by P(r) = 150r/(r+0.5)^2, where r is in ohms. When is the power maximum?
a) 600 ohms
b) 74.5 ohms
c) 0.5 ohms
d) 1.5 ohms
2007-05-30 03:39:26 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The correct answer is (c)
You have to evaluate 150r/(r+0.5)^2 for each of the r (600,74.5,0.5,1.5) and find out the value of P. This can be done by a calculator.
You can see that when r=0.5 ohms, P has the maximum value. I hope I'm right.
2007-05-30 04:13:20 · answer #1 · answered by cidyah 7 · 1⤊ 0⤋
You can find where the power is maximum by setting the first derivative equal to zero.
Hint: If all your questions are multiple choice, then you don't even have to do calculus to find the answer.
Simply plug each answer choice in for r. Whichever one gives you the highest value is the maximum.
2007-05-30 03:43:52 · answer #2 · answered by MsMath 7 · 0⤊ 1⤋
Ans: c) 0.5 ohms
take derivative
then make it equals to zero
r=75/150=0,5
2007-05-30 03:51:08 · answer #3 · answered by iyiogrenci 6 · 1⤊ 0⤋
a. 600
you derive it w.r.t r
and replace r by 0
2007-05-30 03:48:17 · answer #4 · answered by Robin 4 · 1⤊ 0⤋
The answer is C.
2007-05-30 03:44:18 · answer #5 · answered by bruinfan 7 · 1⤊ 0⤋
c
2007-05-30 03:44:32 · answer #6 · answered by rick c 2 · 1⤊ 0⤋