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Find the maximum value of this expression

  1. Jun 12, 2013 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    [itex]\dfrac{tan\left( \theta + 2\pi /3 \right) - tan\left( \theta + \pi /6 \right) + cos\left( \theta + \pi /6 \right)}{\sqrt{3}} \\

    \theta \in \left( -5 \pi /12, - \pi /3 \right)
    [/itex]


    3. The attempt at a solution
    At maximum, f'(x)=0

    sin(θ+∏/6)=0

    But there is no theta in the given interval that satisfies this equation.
     
  2. jcsd
  3. Jun 12, 2013 #2
    I haven't checked whether you did your differentiation correctly (something that you definitely want to check), but in general, for functions defined on a finite domain, the global maximum and minimum points need not occur at stationary points (which correspond to local minima / maxima). The global maximum and minimum can also be attained at the endpoints of the interval.
     
  4. Jun 12, 2013 #3

    CAF123

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    To maximise this expression means you want to maximise the numerator of that expression.
    Consider each term individually and find θ in domain such that each term is ether maximised or minimised depending on the sign preceding it.

    Sketch a graph or note that tan and cos are strictly increasing functions on this domain.
     
  5. Jun 12, 2013 #4

    utkarshakash

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    I later thought about this. Let me try out.
     
  6. Jun 12, 2013 #5
    Instead of graph, I think you can do it algebraically too.

    Rewrite ##\tan(\theta+2\pi/3)=-\cot(\theta+\pi/6)##.
    Let ##\theta+\pi/6=t##. Then ##-\pi/4<t<-\pi/6##.

    Now it's much easier to solve. You can find the derivative or analyze the graph of the resulting expression.

    Hope that helped.
     
  7. Jun 12, 2013 #6

    Ray Vickson

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    Not on an open interval. On an open interval there can be a supremum at an endpoint, but no maximum. Of course, one might speak informally of a maximum in such a case, but it is not technically correct.
     
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