Find the maximum value of this expression

[utkarshakash]

Homework Statement

$\dfrac{tan\left( \theta + 2\pi /3 \right) - tan\left( \theta + \pi /6 \right) + cos\left( \theta + \pi /6 \right)}{\sqrt{3}} \\ \theta \in \left( -5 \pi /12, - \pi /3 \right)$

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.

 

[Fightfish]

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.

 

[CAF123]

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.

 

[utkarshakash]

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.

I later thought about this. Let me try out.

 

[Saitama]

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.

 

[Ray Vickson]

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.

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.