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Homework Help: Finding the minimum

  1. May 22, 2010 #1
    Calculate the minimum of PT = 16.8e0.0697t + 204e-0.356t

    PT' = 0.0697 x 16.8e0.0697t + -.356 x 204e-0.356t

    Stationary points occur when PT'=0. Therefore:

    0=0.0697 x 16.8e0.0697t + -.356 x 204e-0.356t

    My final answer was that t = 4.29. However, the Graphics Calculator begs to differ and says that it is 5.695.

    Attached is my working; where have I gone wrong?
     

    Attached Files:

    Last edited: May 22, 2010
  2. jcsd
  3. May 22, 2010 #2

    cronxeh

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  4. May 22, 2010 #3
    Last edited by a moderator: Apr 25, 2017
  5. May 22, 2010 #4

    cronxeh

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    Gold Member

    Up to this point you were doing well, then you decided to start dividing and made an error. Try by multiplying it out first then dividing:

    0.356*204*exp(-0.356*t) = 0.0697*16.8*exp(0.0697*t)

    It should be

    72.624*exp(-0.356*t) = 1.17096*exp(0.0697*t)
    62.020*exp(-0.356*t)=exp(0.0697*t)

    62.020=exp(0.0697*t)/exp(-0.356*t)

    ln(62.020)=ln(exp(0.4257*t))

    4.1275 = 0.4257*t

    t=9.6957
     
    Last edited: May 22, 2010
  6. May 22, 2010 #5
    In your work, you seem to have changed 0.30 .... into 0.030.... a little below the middle of the page.
     
  7. May 22, 2010 #6
    I did that and now I get 9.695...
     
  8. May 22, 2010 #7
    I rectified that afterwards. Perhaps what I graphed into my Graphics Calculator is wrong.
     
  9. May 22, 2010 #8
    It is. Silly me.
     
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