L in terms of a and b as a maximum

[rebeka]

Homework Statement

$$L = \frac{a}{\sin{\theta}} + \frac{b}{\cos{\theta}}$$find L in terms of a and b where L is a maximum ...

Homework Equations

place long list of trigonometric identities here??

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!

 

[sachinism]

this should help you:

http://everything2.com/title/Answer%3A+Carrying+a+ladder+around+a+corner

 

[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.

 

[rebeka]

Thank you both!

o~o