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Homework Help: Calculus - Max/Min

  1. Apr 1, 2008 #1
    1. The problem statement, all variables and given/known data
    The electric power in Watts produced is given by p= 144r/(r+.8)^2
    where r is the resistance in ohms.
    For what value of r is the power P a maximum?

    2. Relevant equations

    3. The attempt at a solution
    Using the quotient rule i found a derivative of..

    dp/dt= 144(-r+.8) / (r + .8)^3

    Now I need to set this equal to zero and solve for R?

    Multiplying the demonminator by 0 would leave me with just the numerator giving me..
    -144r + 115.2 = 0

    Solving for r gives me r = 0.8ohms

    Was there maybe a problem with my derivative?
  2. jcsd
  3. Apr 1, 2008 #2
    You did it right, find where the derivative is equal to 0 ... [tex]144(.8-r)=0[/tex]

    Solve for r which you did! Is that not the correct answer?
    Last edited: Apr 2, 2008
  4. Apr 2, 2008 #3
    Looks right to me. You can always check to see if you've found a local maximum using the first or second derivative tests and plugging it back into the original function.
  5. Apr 2, 2008 #4
    Test values before and after your max-value: positive to negative should indicate a max-point.
  6. Apr 2, 2008 #5
    Thanks for confirming it for me. Plugging .8 into the derivative gives me 0 so i guess i was right all along.
  7. Apr 2, 2008 #6
    Actually, you'd want to plug in something less than 0.8 and more than 0.8 in the derivative to see if it changes from positive to negative in order for 0.8 to be a maximum.
  8. Apr 2, 2008 #7
    Plot the power function for vallues of r from 0 to 10 using a
    computer algebra system like Mathematica and you will verify that
    the answer 0.8 for the maximum value looks correct.
  9. Apr 2, 2008 #8
    You don't need to do all that! Just do what Snazzy or I suggested. Make good use of your Algebra skills!
  10. Apr 2, 2008 #9
    Yeah, with 1st derivative test right?
    I just figured that it would be ok to assume a maximum because thats what the question asked for, and solving for r only gave me one value.
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