Maximizing Gamma with Lagrange Multipliers

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SUMMARY

This discussion focuses on maximizing the variable gamma using the method of Lagrange multipliers within the context of two constraints: \( g(a,h,r,v,\gamma,\mu)=rv\cos(\gamma)-h=0 \) and \( h(a,h,r,v,\gamma,\mu)=\frac{v^2}{2}-\frac{\mu}{r}+\frac{\mu}{2a}=0 \). The objective function is defined as \( f(a,h,r,v,\gamma,\mu)=\gamma \). Participants clarify the use of multipliers, suggesting that while \( \mu \) is a variable, another multiplier, such as \( \beta \), can be introduced to facilitate the optimization process.

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Given the equations
$$
rv\cos\gamma - h = 0,\quad \frac{v^2}{2} - \frac{\mu}{r} + \frac{\mu}{2a} = 0
$$
I want to maximize gamma.
Do I have to solve for gamma in the first equation to use the method of Lagrange multipliers, or if not, how would I do this in the current form?
 
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Re: Langrange multipliers

If I understand correctly, we have 6 variables, 2 constraints, and the objective function.

The objective function is:

$f(a,h,r,v,\gamma,\mu)=\gamma$

subject to the constraints:

$g(a,h,r,v,\gamma,\mu)=rv\cos(\gamma)-h=0$

$h(a,h,r,v,\gamma,\mu)=\dfrac{v^2}{2}-\dfrac{\mu}{r}+\dfrac{\mu}{2a}=0$

Normally, we use a multiplier for each constraint, and $\lambda$ and $\mu$ are used, but since $\mu$ is one of our variables, would may choose another, such as $\beta$ to be the second multiplier.

So, what we would wind up with is:

$\displaystyle 0=\lambda(0)+\beta\left(-\frac{\mu}{2a^2} \right)$

$\displaystyle 0=\lambda(-1)+\beta(0)$

$\displaystyle 0=\lambda(v\cos(\gamma))+\beta\left(\frac{\mu}{r^2} \right)$

$\displaystyle 0=\lambda(r\cos(\gamma))+\beta(v)$

$\displaystyle 1=\lambda(-rv\sin(\gamma))+\beta(0)$

$\displaystyle 0=\lambda(0)+\beta\left(-\frac{1}{r}+\frac{1}{2a} \right)$

Now, from this system, you need to draw out implications regarding the variables.
 
Re: Langrange multipliers

I found an easier way to do it though.
 
Re: Langrange multipliers

Good, because what I posted did not look like any fun at all! (Yes)
 

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