(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

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:

[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!

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# Homework Help: L in terms of a and b as a maximum

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