Rope between inclines with maximum length of hanging middle portion

[Meden Agan]

And so guys, what shall I do to go on?

 

[Steve4Physics]

And so guys, what shall I do to go on?

You can already solve (I think) the symmetric case where $\alpha = \beta$ and $s_{\alpha} = s_{\beta}$.

What remains is to demonstrate that this symmetry is necessary to achieve the maximum catenary length. Is that what are you trying to achieve? I can't offer any suggestions on this.

 

[Meden Agan]

What remains is to demonstrate that this symmetry is necessary to achieve the maximum catenary length. Is that what are you trying to achieve?

Yes.

 

[Meden Agan]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

 

[haruspex]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

Even if you can, how would you show the third condition is satisfied?

What about my post #21? I haven’t seen a reaction to it since I clarified the idea in post #24.

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$. It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

The total normal force from the slope is $\lambda sg\cos(\alpha)$, and its vertical component is $\lambda sg\cos^2(\alpha)$.

The total upslope friction is $\mu\lambda sg\cos(\alpha)$, and its vertical component is $\mu\lambda sg\sin(\alpha)\cos(\alpha)$.

Adding, $\lambda sg(\cos^2(\alpha)+\mu\sin(\alpha)\cos(\alpha))=\lambda g(s+r)$.

$1+\frac rs=\frac 12(\mu\sin(2\alpha)+1+\cos(2\alpha))$

Maximising wrt $\alpha$: $\mu=\tan(2\alpha)$.

 

[Steve4Physics]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

I don't think it works. E.g. consider @haruspex 's (minimisation) example in Post #8:
$(x-a)^2+(y-b)^2)(x-b)^2+(y-a)^2$

Even when $a \ne b$ the minimum is obtained at two points, $(a,b)$ and $(b,a)$.

Also note that the original problem is a a four (not two) variable problem - all four variables ($\alpha, \beta, s_{\alpha}$ and $s_{\beta}$) must be considered.

EDIT. That's all rendered moot following @haruspex's Post #35 analysis.

 

[Meden Agan]

Even if you can, how would you show the third condition is satisfied?

You're spot on. We could have theoretically proved the third condition rigorously by taking the derivatives with respect to $\alpha$ and $\beta$ and showing that they are equal to zero at a single point $\alpha = \beta$. But, this doesn't happen, as I wrote in post #9.
I suspect this is because, in Multivariable Calculus, taking a partial derivative alone is not sufficient to conclude that we’ve found an extremum.
Maybe we can refer to the second-derivative test for functions of two variables (the Hessian test). However, I fear performing it analytically can be quite tedious.

Let $L_h$ be the length of the hanging section of the cord and $L$ be the total length of the cord. Then, $$\frac{L_h}{L} = \frac{\tan \alpha + \tan \beta}{\dfrac{1}{\cos \alpha \left(\mu \cos \alpha - \sin \alpha \right)} + \tan \alpha + \dfrac{1}{\cos \beta \left(\mu \cos \beta- \sin \beta\right)} + \tan \beta}, \tag{1}$$ as I said in post #15. I sketched an approximate graph following a logically grounded outline, through a heat map of $L_h/L$ vs. angles $\alpha, \beta$ for $\mu = 1$. Here it is the plot with the maximum marked with a white cross.

The function is symmetric (as per $(1)$) and the further the point away from the line of symmetry the lower the value of the function. We take an interval from point infinitely far away from the line to the diagonal itself and say that there is no local maximum. Same applies for other side, means the maximum is on the diagonal itself. So, the maximum is achieved when $\alpha = \beta$.
I fear a more involved analysis using the full three-dimensional slope tends to be cumbersome.

What about my post #21? I haven’t seen a reaction to it since I clarified the idea in post #24.

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$.

At first, I didn't fully understand post #21. If you could make an approximate graphical sketch of your setup, I would write down the required maths :-)

It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

How would you show that?

 

[Steve4Physics]

I’d like to throw this in.

In the diagram below we have P as the lowest point (vertex) of the catenary and lengths
$\text{AB} = s_{\alpha}, \text{arcBP} = r_{\alpha}, \text{arcPC} = r_{\beta}$ and $\text{CD} = s_{\beta}$.

In https://en.wikipedia.org/wiki/Catenary there is a reference to the 'Whewell equation' for the catenary. In the context of the current problem I interpret this as telling us (surprisingly) that arc length and tangential angle are proportional such that:$$\frac {r_{\alpha}}{r_{\beta}} = \frac {\tan \alpha} {\tan \beta}$$So this is an additional relationship which affects the catenary length.

I’m not sure if/how to use this and it’s my bedtime. No doubt I’ll have catenary dreams (or nightmares).

Edits as I can't preview to check LaTeX.
Edit to improve diagram (though it still looks like point P should be a bit more to the left).

 

[haruspex]

Edits as I can't preview to check LaTeX.

I've been doing it by first taking a copy of the post I'm about to make, just as text, then doing preview/refresh in some combination. This shows the preview correctly but it screws the post, so discard the post, make any corrections in the saved copy, then paste that back in and post it.

 

[haruspex]

We could have theoretically proved the third condition rigorously by taking the derivatives with respect to α and β and showing that they are equal to zero at a single point α=β. But, this doesn't happen, as I wrote in post #9.

But the result you got there, with opposite signs, cannot be right. As you agreed, swapping the angles around must convert Exp 1 into Exp 2, so swapping the angles in a solution should also be a solution.

taking a partial derivative alone is not sufficient to conclude that we’ve found an extremum

True, but it largely works. It’s just that instead of having only to discriminate maxima from minima you also have to worry about saddle points.

If you could make an approximate graphical sketch of your setup, I would write down the required maths :-)

@Steve4Physics has kindly done that in post #38, though I would have put point P a bit to the left; doesn’t seem to be at the bottom of the curve (or could be my astigmatism).

How would you show that?

Given an expression of the form $F=\frac{p+q}{r+s}$, and having shown that $\frac{p}{r}\leq t$ and $\frac{q}{s}\leq t$, all variables positive, what bound can you put on $F$?

 

[erobz]

Don't you have the constraint from arc length to somehow use yet?

$$ s_{\alpha} + s_{ \beta} - L = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx $$

 

[haruspex]

Don't you have the constraint from arc length to somehow use yet?

$$ s_{\alpha} + s_{ \beta} - L = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx $$

As in another recent thread on hanging ropes, it does not seem to matter whether the hanging section of the rope is uniform. There are enough equations without it, so it must be irrelevant.

 

[Meden Agan]

Given an expression of the form $F=\frac{p+q}{r+s}$, and having shown that $\frac{p}{r}\leq t$ and $\frac{q}{s}\leq t$, all variables positive, what bound can you put on $F$?

We have $$\frac{p}{r} \le t. \tag 1$$ Multiplying both sides of the inequality by $r > 0$, we have $p \le t r$. By adding $q > 0$ to both sides of the previous inequality, we obtain $$p + q \le tr + q. \tag{1.1}$$

Same way, we have $$\frac{q}{s} \le t. \tag 2$$ Multiplying both sides of the inequality by $s > 0$, we have $q \le t s$. By adding $tr > 0$ to both sides of the previous inequality, we obtain $$q + tr \le ts + tr. \tag{2.1}$$

Taking the two inequalities $(1.1)$ and $(2.1)$, we obtain the chain of inequalities $$p + q \le tr + q \le ts + tr.$$ Taking the first and the last one:
$$\begin{aligned}
p + q &\le ts + tr \\
&= t(s+r),
\end{aligned}$$
or, equivalently,
$$\boxed{\underbrace{\frac{p+q}{s+r}}_{\displaystyle F} \le t}.$$


But, I've a few doubts. Here $t = \left(\dfrac{r}{s}\right)_{\max}$, with $\left(\dfrac{r}{s}\right)_{\max}$ obtained by substituting $\alpha_\max$ for $\alpha$ into the expression for $r/s$. Correct?
If so, haven't we only shown that $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$ is maximized by $\left(\dfrac{r}{s}\right)_{\max}$? It would still remain to show explicitly that $\alpha_\max$ is the angle that also maximizes $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$.

 

[haruspex]

We have $$\frac{p}{r} \le t. \tag 1$$ Multiplying both sides of the inequality by $r > 0$, we have $p \le t r$. By adding $q > 0$ to both sides of the previous inequality, we obtain $$p + q \le tr + q. \tag{1.1}$$

Same way, we have $$\frac{q}{s} \le t. \tag 2$$ Multiplying both sides of the inequality by $s > 0$, we have $q \le t s$. By adding $tr > 0$ to both sides of the previous inequality, we obtain $$q + tr \le ts + tr. \tag{2.1}$$

Taking the two inequalities $(1.1)$ and $(2.1)$, we obtain the chain of inequalities $$p + q \le tr + q \le ts + tr.$$ Taking the first and the last one:
$$\begin{aligned}
p + q &\le ts + tr \\
&= t(s+r),
\end{aligned}$$
or, equivalently,
$$\boxed{\underbrace{\frac{p+q}{s+r}}_{\displaystyle F} \le t}.$$


But, I've a few doubts. Here $t = \left(\dfrac{r}{s}\right)_{\max}$, with $\left(\dfrac{r}{s}\right)_{\max}$ obtained by substituting $\alpha_\max$ for $\alpha$ into the expression for $r/s$. Correct?
If so, haven't we only shown that $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$ is maximized by $\left(\dfrac{r}{s}\right)_{\max}$? It would still remain to show explicitly that $\alpha_\max$ is the angle that also maximizes $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$.

As you found, $F\leq t$, and that does it.
You previously found that $t$ is the maximum value of each of $p/r$ and $q/s$, and you found an angle that achieves that maximum for each.
You also knew that $F$ achieves that value if you plug in that angle for both sides. What was missing was proof that no higher value could be achieved for $F$ by allowing different angles.

 

[Meden Agan]

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$. It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

Here below is my attempt. Please take a look at it and tell me if there are any flaws.

Let us take an infinitesimal piece of cord, long $\mathrm ds$, resting on a plane inclined at an angle $\alpha$ to the horizontal. The weight of the elements is $\lambda g \, \mathrm ds$, where $\lambda$ is the linear density of the cord. The component parallel to the plane (directed downwards) is $\lambda g \sin \alpha \, \mathrm ds$. The normal reaction is $\lambda g \cos \alpha \, \mathrm ds$, so the maximum static friction available (directed upwards) is $\mu \, \lambda g \cos \alpha \, \mathrm ds$. Writing the axial equilibrium (positive upward direction) according to Newton's $2^{\text{nd}}$ Law of motion: $$T(s+ \mathrm ds) - T(s) = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right), \tag{1}$$ and taking the limit $\mathrm ds \to 0$ on the LHS, we have $T(s+ \mathrm ds) - T(s) \to \mathrm dT$; so, $(1)$ becomes $$\mathrm dT = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right). \tag{2}$$ Integrating $(2)$ from the point where the rope detaches from the slope $(T= T_\alpha, \, s = s_\alpha)$ to the top end of the rope $(T_{\text{top}}=0, \, s=0)$ gives $$T_\alpha = \lambda g \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right). \tag{3}$$

Let the length of the hanging section on that side be $r_\alpha$. So, its weight is $\lambda g \, r_\alpha$.
This way, the total length $L_{\text{tot, left}}$ of the cord under consideration is equal to the length of the cord from its left top end to the point where the string detaches from the left slope $(s_\alpha)$, plus the length of the hanging section on the left side $(r_\alpha)$.
Thus, $L_{\text{tot, left}} = s_\alpha + r_\alpha$.

The weight of the left hanging section of the rope, $\lambda g \, r_\alpha$, must be balanced by the vertical component, $T_\alpha \sin \alpha$, of the tension at the point where it joins the part of the cord touching the left incline. We hence have $T_\alpha \sin \alpha = \lambda g \, r_\alpha$ or, equivalently, $$T_\alpha = \frac{\lambda g \, r_\alpha}{\sin \alpha}. \tag{4}$$

Equating the two expressions $(3)$ and $(4)$ for $T_\alpha$:
$$\begin{aligned}
\lambda g \, s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{\lambda g \, r_\alpha}{\sin \alpha} \\
\cancel{\lambda g} \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right) &= \frac{\cancel{\lambda g} \, r_\alpha}{\sin \alpha} \\
s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{r_\alpha}{\sin \alpha} \\
s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right) &= r_\alpha,
\end{aligned}$$
which gives $$r_\alpha = s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right). \tag{5}$$

The suspended fraction $f$ we are trying to maximize is $f = \dfrac{r_\alpha}{L_{\text{tot, left}}}$. So:
$$\begin{aligned}
f &= \frac{r_\alpha}{L_{\text{tot, left}}} \\
&= \frac{r_\alpha}{s_\alpha + r_\alpha} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha + s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\cancel{s_\alpha} \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{\cancel{s_\alpha} \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)},
\end{aligned}$$
thus $f=\dfrac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}$.

Let $F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)$. Then: $$f = \frac{F(\alpha)}{1+ F(\alpha)} \qquad \text{where} \quad F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right).$$
This expression for $f$ clearly is a monotonically increasing function of $F(\alpha)$. And since the suspended fraction increases as $F(\alpha)$ increases, it suffices to maximise $F(\alpha)$ in order to maximise $f$.
Differentiating $F(\alpha)$, we have $F'(\alpha) = \mu \cos(2 \alpha) - \sin(2 \alpha)$. It is zero when $\mu = \tan (2 \alpha_{\max})$. Then: $$\boxed{\alpha_{\max} = \frac{\arctan \mu}{2}},$$ as desired.


Analogous reasoning leads to $\beta_{\max} = \dfrac{\arctan \mu}{2}$, so $\alpha_{\max} = \beta_{\max}$.

 

[Steve4Physics]

Bu@Steve4Physics has kindly done that in post #38, though I would have put point P a bit to the left; doesn’t seem to be at the bottom of the curve (or could be my astigmatism).

Probably my astigmatism!

Point P edited (moved a bit left) in post #38.

 

[haruspex]

Here below is my attempt. Please take a look at it and tell me if there are any flaws.

Let us take an infinitesimal piece of cord, long $\mathrm ds$, resting on a plane inclined at an angle $\alpha$ to the horizontal. The weight of the elements is $\lambda g \, \mathrm ds$, where $\lambda$ is the linear density of the cord. The component parallel to the plane (directed downwards) is $\lambda g \sin \alpha \, \mathrm ds$. The normal reaction is $\lambda g \cos \alpha \, \mathrm ds$, so the maximum static friction available (directed upwards) is $\mu \, \lambda g \cos \alpha \, \mathrm ds$. Writing the axial equilibrium (positive upward direction) according to Newton's $2^{\text{nd}}$ Law of motion: $$T(s+ \mathrm ds) - T(s) = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \theta \alpha \right), \tag{1}$$ and taking the limit $\mathrm ds \to 0$ on the LHS, we have $T(s+ \mathrm ds) - T(s) \to \mathrm dT$; so, $(1)$ becomes $$\mathrm dT = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right). \tag{2}$$ Integrating $(2)$ from the point where the rope detaches from the slope $(T= T_\alpha, \, s = s_\alpha)$ to the top end of the rope $(T_{\text{top}}=0, \, s=0)$ gives $$T_\alpha = \lambda g \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right). \tag{3}$$

Let the length of the hanging section on that side be $r_\alpha$. So, its weight is $\lambda g \, r_\alpha$.
This way, the total length $L_{\text{tot, left}}$ of the cord under consideration is equal to the length of the cord from its left top end to the point where the string detaches from the left slope $(s_\alpha)$, plus the length of the hanging section on the left side $(r_\alpha)$.
Thus, $L_{\text{tot, left}} = s_\alpha + r_\alpha$.

The weight of the left hanging section of the rope, $\lambda g \, r_\alpha$, must be balanced by the vertical component, $T_\alpha \sin \alpha$, of the tension at the point where it joins the part of the cord touching the left incline. We hence have $T_\alpha \sin \alpha = \lambda g \, r_\alpha$ or, equivalently, $$T_\alpha = \frac{\lambda g \, r_\alpha}{\sin \alpha}. \tag{4}$$

Equating the two expressions $(3)$ and $(4)$ for $T_\alpha$:
$$\begin{aligned}
\lambda g \, s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{\lambda g \, r_\alpha}{\sin \alpha} \\
\cancel{\lambda g} \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right) &= \frac{\cancel{\lambda g} \, r_\alpha}{\sin \alpha} \\
s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{r_\alpha}{\sin \alpha} \\
s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right) &= r_\alpha,
\end{aligned}$$
which gives $$r_\alpha = s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right). \tag{5}$$

The suspended fraction $f$ we are trying to maximize is $f = \dfrac{r_\alpha}{L_{\text{tot, left}}}$. So:
$$\begin{aligned}
f &= \frac{r_\alpha}{L_{\text{tot, left}}} \\
&= \frac{r_\alpha}{s_\alpha + r_\alpha} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha + s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\cancel{s_\alpha} \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{\cancel{s_\alpha} \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)},
\end{aligned}$$
thus $f=\dfrac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}$.

Let $F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)$. Then: $$f = \frac{F(\alpha)}{1+ F(\alpha)} \qquad \text{where} \quad F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right).$$
This expression for $f$ clearly is a monotonically increasing function of $F(\alpha)$. And since the suspended fraction increases as $F(\alpha)$ increases, it suffices to maximise $F(\alpha)$ in order to maximise $f$.
Differentiating $F(\alpha)$, we have $F'(\alpha) = \mu \cos(2 \alpha) - \sin(2 \alpha)$. It is zero when $\mu = \tan (2 \alpha_{\max})$. Then: $$\boxed{\alpha_{\max} = \frac{\arctan \mu}{2}},$$ as desired.


Analogous reasoning leads to $\beta_{\max} = \dfrac{\arctan \mu}{2}$, so $\alpha_{\max} = \beta_{\max}$.

All looks ok to me.