Maximization problem — Stiffest beam that can be cut from a log

[Karol]

Again quick glance. Something may be right and something must be wrong. You expect the answer, S to be a like length to the fourth power and that isn't.
9¼ is √3.

$$\eta=\frac{1}{3}~\rightarrow~\sqrt[4]{a^2 (\eta b^2)(\eta b^2)(\eta b^2)}=\sqrt[4]{a^2\frac{1}{9}(b^3)^2}=\frac{1}{\sqrt[4]{9}}\sqrt[4]{(ab^3)^2}=\frac{1}{\sqrt{3}}\sqrt{ab^3} \leq \frac{r^2}{4}$$
$$\sqrt{ab^3} \leq \frac{\sqrt{3}}{4}r^2~~\Rightarrow~~ab^3\leq\frac{3}{16}r^4$$
$$a\sqrt{(r^2-a^2)^3}\leq\frac{3}{16}r^4$$
The units of both sides are $~[m]^4~$. if i will succeed to isolate a, which i can't, it will be O.K. since a will have units of [m]

Ex 5:
$$\left\{ \begin{array}{l} S=a(4r^2-a^2)^{3/2} \\ a=r \end{array} \right. ~\Rightarrow~S=\sqrt{27}r^4$$
I use ≤ in order to organize my calculations. i am still confused. this tells me what is the right side and what is left.
S(a)≤f(r), and the biggest S is when S(a)=f(r)

 

[StoneTemplePython]

$$\eta=\frac{1}{3}~\rightarrow~\sqrt[4]{a^2 (\eta b^2)(\eta b^2)(\eta b^2)}=\sqrt[4]{a^2\frac{1}{9}(b^3)^2} \leq\frac{a^2+3\cdot \frac{1}{3}b^2}{4}=\frac{a^2+b^2}{4}$$
$$\frac{1}{\sqrt[4]{9}}\sqrt{a^2(b^3)^2}=\frac{1}{\sqrt[4]{9}}\sqrt{S^2}=\frac{1}{\sqrt[4]{9}}\cdot S\leq \frac{r^2}{4}$$
$$S=a\big( \sqrt{r^2-a^2} \big) \leq\ \frac{\sqrt[4]{9}}{4}r^2$$
Can't isolate $~a$

ok $\eta = 3$ is correct. Remember $\eta = \gamma^\frac{1}{3}$.

Now go back to your earlier posting:

If i could find a $~\gamma~$ such that $~\gamma a^2=b^2~$ then there would be equality

you should be able to 'isolate' one of these (i.e. convert from one to the other, given these equality conditions) with this information. Or what I said on the prior page:

To put it a different way: you are interested maximizing the left hand side, so you want to find a $\gamma$ that allows you to write the above inequality, and you know the upper bound is achievable if and only if$x_1 = x_2 = x_3 = x_4$. The upper bound is written purely in terms of this fixed radius value that the problem gives you, so it is a "good" upper bound.

I'm not sure what else can be said here. You have the scalar needed and you know the conditions under which the equality can be achieved. You just need to put them together.

- - - -
edit: typo. It should say $\eta = \frac{1}{3}$

 

[epenguin]

Not checked all working but watch out – you still seem in two minds whether you want to use (r² - a²) or (4r² - a²). I think it still needs some checking. I am now making mistakes myself and I have to travel.

I think I haven't given you the right exercise, just the right type of exercise. It is hard to see in the mouth just just what you think the optimum a is. It cannot be r . If a Is the half-Width, that corresponds to collapse of the rectangle into just a line, the weakest structure a minimum that might come out of your calculation, and if a Is the total with, then it Is impossible it equal r. I think the whole thing could benefit from being written out againFrom start to conclusion; if you get it right it is quite simple just a few lines, and would actually save you and everyone time.

When you say you cannot isolate a, When you have got a and b in terms of r, that is all you need - gives you the strongest shape

 

[Karol]

and if a Is the total with, then it Is impossible it equal r

I drew a=r:

ok $~\eta = 3~$ is correct. Remember $~\eta = \gamma^\frac{1}{3}~$.

I found $\eta=\frac{1}{3}~$, not 3.
The right side isn't important. it's not important what the stiffness is as long as it's the maximum.
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3b^2=r^2~\rightarrow~a=\frac{r}{2}$$
Wrong. in this answer a is the whole width

 

[StoneTemplePython]

I drew a=r:
View attachment 221572

I found $\eta=\frac{1}{3}~$, not 3.
The right side isn't important. it's not important what the stiffness is as long as it's the maximum.
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3b^2=r^2~\rightarrow~a=\frac{r}{2}$$
Wrong. in this answer a is the whole width

Yes -- typo on my part about $\eta$ as I was running to the airport. I edited the post.

The relation is clearly that to reach the upper bound you must have the case that $a^2 = \eta b^2 = \frac{1}{3} b^2$
taking advantage of positivity, this gives

$a = \frac{1}{\sqrt{3}}b$

- - - -
I am not sure what you did here otherwise. If $a$ "is the whole width"-- I'm not sure what that means... If you look at my post #5 I said

you have $\text{payoff function} \propto ab^3$ and a constraint $\sqrt{a^2 + b^2} = r$ or something along those lines

you'll have to figure out that mapping between my verbiage and the actual problem. Re-visiting the drawing, it looks like the most direct interpretation is actually

$\big(\frac{1}{2}a\big)^2 + \big(\frac{1}{2}b\big)^2 = r^2 = \frac{1}{4}a^2+ \frac{1}{4}b^2$

but this 'refinement' is mere rescaling, and irrelevant given that your problem statement doesn't actually give equations -- it just gives a $\propto$ relation. Tracing through the equality conditions is really what's important.

In any case this is sufficiently subtle that I'll re-visit after some rest.

 

[Karol]

I made a small mistake, again:
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3a^2=r^2~\rightarrow~a=\frac{r}{2}$$
$a~$ should equal $~r$

 

[StoneTemplePython]

I made a small mistake, again:
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3a^2=r^2~\rightarrow~a=\frac{r}{2}$$
$a~$ should equal $~r$

I you want to be very literal about this, consider that, after re-examining the picture, what we actually have as our constraint is that

$a^2 + b^2 = d^2$ where $d$ is the diameter -- I was being loose by using $r$ (though it works up to a rescaling) but, again, the literal interpretation is to use $d$ as the diameter for the constraint. None of the inequality work changes after substituting in $d$ in place of $r$. Using your work, but with the substituted value, you get:

$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=d^2~\rightarrow~a^2+3a^2=d^2~\rightarrow~a=\frac{d}{2} = r$$

 

[epenguin]

I drew a=r:
View attachment 221572

I found $\eta=\frac{1}{3}~$, not 3.
The right side isn't important. it's not important what the stiffness is as long as it's the maximum.
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3b^2=r^2~\rightarrow~a=\frac{r}{2}$$
Wrong. in this answer a is the whole width

OK The whole with can equal the radius.And the answer to this problem is that it does. The radius and the base form an equilateral triangle. Trouble is you seem seem never to have made up your mind whether a and b represent sides or half sides of the rectangle. In this very post you seem to use both meanings. I don't know what this whole thread is about, except confusion caused by this, certainly all these γ's and η's are unnecessary. Let $x, y$ mean half side, then
$$S=16kxy^3$$
I will just find the $x$ that gives a maximum of $xy^3$ which I call
$$s=xy^3\\s=x\left( r^{2}-x^{2}\right) ^{3/2}\\
s^\prime=\left( r^{2}-x^{2}\right) ^{3/2}-3x^{2}\left( r^{2}-x^{2}\right) ^{1/2}\\
=\left( r^{2}-x^{2}\right) ^{1/2}\left( r^{2}-4x^{2}\right)
$$

So the maximum is where $x=r/2$, ...

I get S_max = 3√3$~r^4$

 

[Karol]

I thank you:
Math_QED
phinds
Orodruin
StoneTemplePhyton
epenguin
Ray Vickson