L in terms of a and b as a maximum

1. Aug 31, 2010

rebeka

1. The problem statement, all variables and given/known data

$$L = \frac{a}{\sin{\theta}} + \frac{b}{\cos{\theta}}$$

find L in terms of a and b where L is a maximum .....

2. Relevant equations

place long list of trigonometric identities here??

3. The attempt at a solution

so my attempt looks something like this:

as a cause of the chain rule:

$$\frac{dL}{d\theta} = - \frac{a \cdot \cos{\theta} }{\sin^2{\theta}} + \frac{b \cdot \sin{\theta}}{\cos^2{\theta}}$$

so with $$\frac{dL}{d\theta} = 0$$ ..

$$\sin{\theta} = \cos{\theta}(\frac{a}{b})^{\frac{1}{3}}$$

$$\cos{\theta} = \frac{\sin{\theta}}{ (\frac{a}{b})^{\frac{1}{3}}}$$

also..

$$L = a\csc{\theta} + b\sec{\theta}$$

so that by identities..

$$\frac{dL}{d \theta} = a(-\cot{\theta} \cdot \csc{\theta}) + b(-\csc^2{\theta})$$

which when $$\frac{dL}{d \theta}$$ is set to zero and the resultant reduced ....

$$\cos{\theta} = - \frac{b}{a}$$

substituting the above into the original equation and reducing I have come to

$$= - \frac{a^{\frac{5}{3}}}{b^{\frac{2}{3}}} - \frac{a \cdot b^{\frac{2}{3}}}{b^{\frac{2}{3}}}$$

which looks like it is approaching the final answer of

$$L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$$

but is most likely somehow wrong and there is probably a much simpler answer I am overlooking as this seems a bit out of context :/

if I skipped too many steps forgive me I will add more .... latex is very time consuming for me!

Last edited: Aug 31, 2010
2. Sep 1, 2010

3. Sep 1, 2010

Dick

You've got tan(theta)=(a/b)^(1/3). Stop there. So theta=arctan((a/b)^(1/3)). Now you need to find sin and cos of the arctan. BTW this isn't necessarily a maximum - so far it's just a critical point.

4. Sep 1, 2010

rebeka

Thank you both!

o~o