L in terms of a and b as a maximum

[rebeka]

Homework Statement

[tex] L = \frac{a}{\sin{\theta}} + \frac{b}{\cos{\theta}} [/tex]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:

[tex] \frac{dL}{d\theta} = - \frac{a \cdot \cos{\theta} }{\sin^2{\theta}} + \frac{b \cdot \sin{\theta}}{\cos^2{\theta}} [/tex]so with [tex] \frac{dL}{d\theta} = 0 [/tex] ..

[tex] \sin{\theta} = \cos{\theta}(\frac{a}{b})^{\frac{1}{3}} [/tex][tex] \cos{\theta} = \frac{\sin{\theta}}{ (\frac{a}{b})^{\frac{1}{3}}} [/tex]also..

[tex] L = a\csc{\theta} + b\sec{\theta} [/tex]so that by identities..

[tex] \frac{dL}{d \theta} = a(-\cot{\theta} \cdot \csc{\theta}) + b(-\csc^2{\theta}) [/tex]which when [tex] \frac{dL}{d \theta} [/tex] is set to zero and the resultant reduced ...

[tex] \cos{\theta} = - \frac{b}{a} [/tex]substituting the above into the original equation and reducing I have come to

[tex] = - \frac{a^{\frac{5}{3}}}{b^{\frac{2}{3}}} - \frac{a \cdot b^{\frac{2}{3}}}{b^{\frac{2}{3}}} [/tex]which looks like it is approaching the final answer of

[tex] L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}} [/tex]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