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

[Karol]

Homework Statement

Homework Equations

Pitagora's: $~a^2+b^2=c^2$
Maxima/minima are where the first derivative is 0

The Attempt at a Solution

$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=\frac{16r^2}{a^2}$$
The strength S has the proportion coefficient k: $~S=kab^2$
$$s=kab^2=k\frac{16r^2}{a^2}$$
$$S'=-16kr^2\dfrac{1}{a^2}$$
I can't equal this to 0

 

[member 587159]

Homework Statement

View attachment 221288

Homework Equations

Pitagora's: $~a^2+b^2=c^2$
Maxima/minima are where the first derivative is 0

The Attempt at a Solution

View attachment 221289
$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=\frac{16r^2}{a^2}$$
The strength S has the proportion coefficient k: $~S=kab^2$
$$s=kab^2=k\frac{16r^2}{a^2}$$
$$S'=-16kr^2\dfrac{1}{a^2}$$
I can't equal this to 0

I find your spelling of Pythagoras most disturbing.

 

[phinds]

The strength S has the proportion coefficient k: $~S=kab^2$

Your might want to reread the problem statement.

 

[Orodruin]

$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=\frac{16r^2}{a^2}$$

This is not correct. The last equality does not follow from the first.

I find your spelling of Pythagoras most disturbing.

Pythagoras is spelled Pitágoras in Spanish. It is a transcription anyway. It should really be written Πυθαγόρας but who has the willpower to write that every time?

 

[StoneTemplePython]

My take:

you have $\text{payoff function} \propto ab^3$ and a constraint $\sqrt{a^2 + b^2} = r$ or something along those lines. If you want to make life easier, take advantage of positivity and look at both the squared payoff function and the squared constraint

$\text{squared payoff function} \propto a^2 \big(b^2\big)^3$

and a squared constraint $a^2 + b^2 = r^2$

Fundamentally you are looking at multiplying positive numbers and relating that product to the the sum of said numbers. This screams $GM \leq AM$. It takes a little modeling insight, but can be easily solved with such an approach.

 

[Karol]

My take:

you have $\text{payoff function} \propto ab^3$ and a constraint $\sqrt{a^2 + b^2} = r$ or something along those lines. If you want to make life easier, take advantage of positivity and look at both the squared payoff function and the squared constraint

$\text{squared payoff function} \propto a^2 \big(b^2\big)^3$

and a squared constraint $a^2 + b^2 = r^2$

Fundamentally you are looking at multiplying positive numbers and relating that product to the the sum of said numbers. This screams $GM \leq AM$. It takes a little modeling insight, but can be easily solved with such an approach.

I still don't know what to do but:
$$\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2}$$
But $~\sqrt{a^2\cdot b^2} < \sqrt{a^2\cdot (b^2)^3}~$ so i can't make $~\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2} > \sqrt{a^2\cdot (b^2)^3}$

How do you make the character marked with an arrow in

?
$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=4r^2-a^2$$
$$F=\alpha ab^3,~~F'=\alpha (b^3+3ab^2)=\alpha [(4r^2-a^2)^{3/2}+3(4r^2-a^2)]$$
$$F'=0~\rightarrow~(4r^2-a^2)^{3/2}=-3a(4r^2-a^2)$$
Can't solve.

 

[Orodruin]

Most important things first.

How do you make the character marked with an arrow in View attachment 221314 ?

This depends on your keyboard settings. I imagine this should be rather straight-forward on a Spanish keyboard. See this page.

Edit: Also note that the language of this forum is English and Pythagoras is spelled Pythagoras in English.

$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=4r^2-a^2$$

Yes, this let's you remove $b$ from your equations in favour of $a$. Note that $a$ and $b$ are not independent parameters. You can express one in terms of the other and so you can express $ab^3$ as a function of one of them only.

$$F=\alpha ab^3,~~F'=\alpha (b^3+3ab^2)=\alpha [(4r^2-a^2)^{3/2}+3(4r^2-a^2)]$$

This is not correct. What are you differentiating with respect to? It seems as if you are trying to treat the derivative as a derivative with respect to both $a$ and $b$ at the same time. Regardless of what you are differentiating with respect to, you cannot have both $a' = 1$ and $b'=1$ at the same time. In particular, note that differentiating the relation between $a$ and $b$ gives you $a \, da + b\, db = 0$.

$$F'=0~\rightarrow~(4r^2-a^2)^{3/2}=-3a(4r^2-a^2)$$
Can't solve.

Since your derivative was wrong, this also means that this is not the correct expression to solve.

I suggest you replace $b$ with its expression as a function of $a$ and then differentiate your expression with respect to $a$ (your target function will be a function of $a$ only).

 

[StoneTemplePython]

I still don't know what to do but:
$$\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2}$$
But $~\sqrt{a^2\cdot b^2} < \sqrt{a^2\cdot (b^2)^3}~$ so i can't make $~\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2} > \sqrt{a^2\cdot (b^2)^3}$

$GM \leq AM$ is a lot more general than 2 terms. One rather general formulation gives you:

$\big(\prod_{k=1}^n x_k\big)^\frac{1}{n} \leq \frac{1}{n} \sum_{k=1}^n x_k$
where each $x_k \geq 0$, and there is equality iff $x_1 = x_2 = ... = x_n$. There are lots of good materials on it. Perhaps this link will help:

https://brilliant.org/wiki/arithmetic-mean-geometric-mean/#arithmetic-mean-geometric-mean
- - - - -

A naive application of $GM\leq AM$ to the problem would give (note: I dropped $\propto$ in favor of $=$ for convenience, below)

$\big(\text{squared payoff function}\big)^\frac{1}{4} = \big(a^2 \big(b^2\big)^3\big)^\frac{1}{4} = \big(a^2 b^2 b^2 b^2\big)^\frac{1}{4} \leq \frac{1}{4}\big(a^2 + b^2 + b^2 + b^2\big) = \frac{1}{4}\big(a^2 + 3b^2 \big)$

which is a failure -- but where and how it fails is instructive . What you want is something like

$\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4} \leq \frac{1}{4}\big(a^2 + b^2 \big) = \frac{1}{4}r^2$

I'm not sure what else I can say without giving it away. If you can figure out that positive scalar $\gamma$, and what it implies for $x_1$ vs $x_2$ vs $x_3$ vs $x_4$, then you very quickly are led to your answer.

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.

 

[Orodruin]

Pitagora is how Πυθαγόρας is rendered in some languages.

As stated in #4 ...

 

[epenguin]

Sorry , removed.

 

[Orodruin]

Sorry , removed.

Don't think there was a need for that. I just found it funny that we probably copied Πυθαγόρας from the same Wikipedia page. (At least I did not have the patience to type it out ...)

 

[epenguin]

Just from symmetry I think you can state straight off and without any calculations what an answer could be, a conjecture at least, then some checking would be advisable.

 

[epenguin]

Don't think there was a need for that. I just found it funny that we probably copied Πυθαγόρας from the same Wikipedia page. (At least I did not have the patience to type it out ...)

Nor me, I always try to go for simplicity - see #12.

 

[epenguin]

Just from symmetry I think you can state straight off and without any calculations what an answer could be, a conjecture at least, then some checking would be advisable.

This sounded like a sound idea but after wasting(?) time on it I decided it isn't.

 

[Ray Vickson]

I still don't know what to do but:
$$\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2}$$
But $~\sqrt{a^2\cdot b^2} < \sqrt{a^2\cdot (b^2)^3}~$ so i can't make $~\frac{a^2+b^2}{2}\geqslant \sqrt{a^2\cdot b^2}~\rightarrow~a^2+b^2\geqslant 2 \sqrt{a^2\cdot b^2} > \sqrt{a^2\cdot (b^2)^3}$

How do you make the character marked with an arrow in View attachment 221314 ?
$$\left( \frac{a}{2} \right)^2+\left( \frac{b}{2} \right)^2=r^2~\rightarrow~b^2=4r^2-a^2$$
$$F=\alpha ab^3,~~F'=\alpha (b^3+3ab^2)=\alpha [(4r^2-a^2)^{3/2}+3(4r^2-a^2)]$$
$$F'=0~\rightarrow~(4r^2-a^2)^{3/2}=-3a(4r^2-a^2)$$
Can't solve.

Use $a = \sqrt{r^2-b^2}$ to write the stiffness $S$ as
$$S = a b^3 = \sqrt{r^2-b^2}\, b^3 = f(b).$$
You want to maximize $f(b)$.

If you are careful, setting $f'(b) = 0$ will give you a problem that can be solved in about two lines of easy work; if you are careless and make errors you will never solve it.

Please do not make the problem harder than it needs to be: just take care!

 

[Karol]

$GM \leq AM$ is a lot more general than 2 terms. One rather general formulation gives you:

$\big(\prod_{k=1}^n x_k\big)^\frac{1}{n} \leq \frac{1}{n} \sum_{k=1}^n x_k$
where each $x_k \geq 0$, and there is equality iff $x_1 = x_2 = ... = x_n$.

A naive application of $GM\leq AM$ to the problem would give (note: I dropped $\propto$ in favor of $=$ for convenience, below)

$\big(\text{squared payoff function}\big)^\frac{1}{4} = \big(a^2 \big(b^2\big)^3\big)^\frac{1}{4} = \big(a^2 b^2 b^2 b^2\big)^\frac{1}{4} \leq \frac{1}{4}\big(a^2 + b^2 + b^2 + b^2\big) = \frac{1}{4}\big(a^2 + 3b^2 \big)$

which is a failure -- but where and how it fails is instructive . What you want is something like

$\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4} \leq \frac{1}{4}\big(a^2 + b^2 \big) = \frac{1}{4}r^2$

I'm not sure what else I can say without giving it away. If you can figure out that positive scalar $\gamma$, and what it implies for $x_1$ vs $x_2$ vs $x_3$ vs $x_4$, then you very quickly are led to your answer.

There are 5 terms in $~\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4}~$ and the root is only 4'th degree (is it correct how i call a $~\sqrt[4]{x}~$ root?)
And what's more, on the right side: $~\frac{1}{4}\big(a^2 + b^2 \big)~$ there are only 2 terms: one $~a^2~$ and one $~b^2$
If i could find a $~\gamma~$ such that $~\gamma a^2=b^2~$ then there would be equality:
$$\sqrt[4]{(\gamma a^2)b^2b^2b^2}=\sqrt[4]{(b^2)^4}=\frac{4 b^2}{4}=b^2$$
But there are 4 terms on the right instead of 2 in the post, and it also doesn't help.

This is not correct. What are you differentiating with respect to? It seems as if you are trying to treat the derivative as a derivative with respect to both $~a~$ and $~b~$ at the same time. Regardless of what you are differentiating with respect to, you cannot have both $~a′=1~$ and $~b′=1~$ at the same time. In particular, note that differentiating the relation between $~a~$ and $~b~$ gives you $~ada+bdb=0~$

If i have only a and only b, why isn't correct that a'=1 and b'=1?
Which relation do you mean between $~a~$ and $~b~$ that gives $~ada+bdb=0~$? $~(ab)'=adb+bda$

My answer:
$$S=ab^3=a\sqrt{r^2-a^2},~S'=\sqrt{r^2-a^2}+a(r^2-a^2)^{-1/2}(-2a)=\sqrt{r^2-a^2}-\frac{2a^2}{\sqrt{r^2-a^2}}$$
$$S'=0~\rightarrow~a=\frac{r}{\sqrt{3}}$$
The answer in the book that the width of the beam is the radius of the log

 

[StoneTemplePython]

There are 5 terms in $~\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4}~$ and the root is only 4'th degree (is it correct how i call a $~\sqrt[4]{x}~$ root?)
And what's more, on the right side: $~\frac{1}{4}\big(a^2 + b^2 \big)~$ there are only 2 terms: one $~a^2~$ and one $~b^2$
If i could find a $~\gamma~$ such that $~\gamma a^2=b^2~$ then there would be equality:
$$\sqrt[4]{(\gamma a^2)b^2b^2b^2}=\sqrt[4]{(b^2)^4}=\frac{4 b^2}{4}=b^2$$
But there are 4 terms on the right instead of 2 in the post, and it also doesn't help.

This is very close. The way to think about it is there are only 4 terms on the LHS. $\gamma$ is just some positive scalar that is there to help you out. What if, instead of

$\sqrt[4]{(\gamma a^2)b^2b^2b^2}$,

you tried the other approach of

$\sqrt[4]{ a^2 (\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)}$

could you make that work?

edit:
if $\gamma^\frac{1}{3}$ is distracting / confusing, you could instead use

$\eta:= \gamma^\frac{1}{3}$, where again $\eta$ is just some positive scalar there to help you out. Written this way, it'd be:

$\big(\eta^3 \text{squared payoff function}\big)^\frac{1}{4} = \sqrt[4]{ a^2 (\eta b^2)(\eta b^2)(\eta b^2)} = \Big( a^2 (\eta b^2)(\eta b^2)(\eta b^2)\Big)^\frac{1}{4}$

- - -

$~\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4}~$ and the root is only 4'th degree (is it correct how i call a $~\sqrt[4]{x}~$ root?)

It's ok to write it this way, though I prefer to show the fraction in the exponent. Why? Because the linkage of $GM \leq AM$ allows you to 'bring down' the fractions in the exponents of a product and make them scalars of a sum, so I find it easier to visualize this way. (These fraction need to behave like probabilities -- real non-negative and sum to one). That's my taste in the matter I suppose.

 

[Orodruin]

If i have only a and only b, why isn't correct that a'=1 and b'=1?

Because $a$ and $b$ are related, which is something you must understand to solve this problem. You cannot arbitrarily change $a$ and $b$ by the same amount.

Which relation do you mean between $~a~$ and $~b~$ that gives $~ada+bdb=0~$? $~(ab)'=adb+bda$

No, that makes no mathematical sense. What you need to differentiate is the relation that gives the relation between $a$ and $b$, i.e., the equation of the circle.

 

[Ray Vickson]

There are 5 terms in $~\big(\gamma* \text{squared payoff function}\big)^\frac{1}{4}~$ and the root is only 4'th degree (is it correct how i call a $~\sqrt[4]{x}~$ root?)
And what's more, on the right side: $~\frac{1}{4}\big(a^2 + b^2 \big)~$ there are only 2 terms: one $~a^2~$ and one $~b^2$
If i could find a $~\gamma~$ such that $~\gamma a^2=b^2~$ then there would be equality:
$$\sqrt[4]{(\gamma a^2)b^2b^2b^2}=\sqrt[4]{(b^2)^4}=\frac{4 b^2}{4}=b^2$$
But there are 4 terms on the right instead of 2 in the post, and it also doesn't help.If i have only a and only b, why isn't correct that a'=1 and b'=1?
Which relation do you mean between $~a~$ and $~b~$ that gives $~ada+bdb=0~$? $~(ab)'=adb+bda$

My answer:
$$S=ab^3=a\sqrt{r^2-a^2},~S'=\sqrt{r^2-a^2}+a(r^2-a^2)^{-1/2}(-2a)=\sqrt{r^2-a^2}-\frac{2a^2}{\sqrt{r^2-a^2}}$$
$$S'=0~\rightarrow~a=\frac{r}{\sqrt{3}}$$
The answer in the book that the width of the beam is the radius of the log

No: you are still not being careful!

We can use the constraint to turn $S = a b^3$ into a univariate function as
$$\begin{array}{rll}S = &\sqrt{r^2-b^2}\, b^3 &\Leftarrow \; \text{TRUE} \\
S =& a \, (r^2-a^2)^{3/2} & \Leftarrow \; \text{TRUE}\\
S =& a \, \sqrt{r^2-a^2} & \Leftarrow \; \text{FALSE!}
\end{array}
$$

 

[Orodruin]

No: you are still not being careful!

I would like to emphasise this even more so I added the italic style ...

Taking care when you do your arithmetics to understand every step that you are doing and that you get it correctly is absolutely fundamental in mathematics and science.

 

[Karol]

$$S=ab^3,~~a^2+b^2=r^2~~\rightarrow~S=a(4r^2-a^2)^{3/2}$$
$$S'=...=4\sqrt{4r^2-a^2}(r^2-a^2)$$
$$S'=0:~\left\{ \begin{array}{lll}r^2=a^2 \Rightarrow & a=r & \text{True} \\ 4r^2=a^2 \Rightarrow & a=2r & \text{False} \end{array} \right.$$

I try to use $~\sqrt[4]{ a^2 (\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)}~$.
$$b^2=4r^2-a^2=(2r+a)(2r-a)$$
No scalar (like3,7,... etc.) $~\gamma^{1/3}~$ can change that to $~a^2~$ alone.
And even if i found, the right side would be $~\frac{4a^2}{4}=a^2~$ and it's not

 

[epenguin]

Maybe contrary contrary to what I said before now cancelled, your differentiation is OK. (I'm just following a slightly different formulation* and have little time to check.)

So assuming it is OK up to a=r, after that what is the problem? You have a straightforward expression for b In terms of a and r, so easy to get b in terms of r.* I think it makes it easier if you make r = 1.
I don't think the optimum shape depends on the size.
Or maybe more to the point, it doesn't depend on the units you use for measuring length. (Get the shape right first and worry about this when you have to calculate the final S).
And I avoid confusing numbers by using instead of rectangle sides, half sides say x=a/2, y=b/2. Then S = 16xy (if I am not mistaken! )

 

[Ray Vickson]

$$S=ab^3,~~a^2+b^2=r^2~~\rightarrow~S=a(4r^2-a^2)^{3/2}$$
$$S'=...=4\sqrt{4r^2-a^2}(r^2-a^2)$$
$$S'=0:~\left\{ \begin{array}{lll}r^2=a^2 \Rightarrow & a=r & \text{True} \\ 4r^2=a^2 \Rightarrow & a=2r & \text{False} \end{array} \right.$$

I try to use $~\sqrt[4]{ a^2 (\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)}~$.
$$b^2=4r^2-a^2=(2r+a)(2r-a)$$
No scalar (like3,7,... etc.) $~\gamma^{1/3}~$ can change that to $~a^2~$ alone.
And even if i found, the right side would be $~\frac{4a^2}{4}=a^2~$ and it's not

How do you go from $a^2+b^2-r^2$ to $b = \sqrt{4r^2-a^2}$ (that is, $b^3 = (4r^2-a^2)^{3/4}$)?

I said it before and I will say it again: you MUST BE MORE CAREFUL! There is absolutely no acceptable excuse for the type of error you just made above (among many others in other posts). I am no longer willing to help.

 

[epenguin]

In this case maybe there is a bit of a mix up - a mistake would be in the first line a² + b² = r² - that should be (2r)² = 4r². My suggestion in my last post is my way by which I reduce but do not eliminate my errors, hopefully additional tricks and signs eliminate most of them.

 

[Karol]

Originally i denoted the width and height as a and b, see my drawing.
Later you,Ray, passed to denote width and height as 2a and 2b in post #15, since you wrote $~a = \sqrt{r^2-b^2}$

I returned, in post #21, to my original notation, i didn't make a mistake. please continue to help.

I didn't explain, in post #15, that after i found the correct answer as a=r, i continued to try to solve in StoneTemple's method GM<AM, post #5, and still cannot solve. that is what i explained in my last post. sorry for not explaining, i hurry to reply for you all not to loose contact and don't explain enough.

 

[StoneTemplePython]

I try to use $~\sqrt[4]{ a^2 (\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)(\gamma^\frac{1}{3} b^2)}~$.

No scalar (like3,7,... etc.) $~\gamma^{1/3}~$ can change that to $~a^2~$ alone.
And even if i found, the right side would be $~\frac{4a^2}{4}=a^2~$ and it's not

I think you can do better. Again, the problem, with somewhat simplified notation is:

$\big(\eta^3 \text{squared payoff function}\big)^\frac{1}{4} = \Big( a^2 (\eta b^2)(\eta b^2)(\eta b^2)\Big)^\frac{1}{4}$

apply $GM\leq AM$

$\big(\eta^3 \text{squared payoff function}\big)^\frac{1}{4} = \Big( a^2 (\eta b^2)(\eta b^2)(\eta b^2)\Big)^\frac{1}{4} \leq \frac{1}{4}\big(a^2 + \eta b^2 + \eta b^2 + \eta b^2\big) = \frac{1}{4}\big(a^2 + 3\eta b^2\big)$

and the goal (from post #8) is to set the Right Hand Side equal to

$\frac{1}{4}\big(a^2 + b^2 \big) = \frac{1}{4}r^2$
- - - -
please tell me you can find some $\eta$ whereby

$\frac{1}{4}\big(a^2 + 3\eta b^2\big) = \frac{1}{4}\big(a^2 + b^2 \big) = \frac{1}{4}r^2$

after finding said $\eta$ you just need to trace back the equality conditions that allow you to reach this upper bound

 

[epenguin]

Originally i denoted the width and height as a and b, see my drawing.
Later you,Ray, passed to denote width and height as 2a and 2b in post #15, since you wrote $~a = \sqrt{r^2-b^2}$

I returned, in post #21, to my original notation, i didn't make a mistake. please continue to help.

I didn't explain, in post #15, that after i found the correct answer as a=r, i continued to try to solve in StoneTemple's method GM<AM, post #5, and still cannot solve. that is what i explained in my last post. sorry for not explaining, i hurry to reply for you all not to loose contact and don't explain enough.

I have only time for skimming these posts, so I can be wrong. I'm fairly sure though that in #21 you have mixed two notations, "ours" and "yours" which is that of the original problem. I think the last equation In the first line of #21 is true in your notation, however then the → which annoyed Ray Is not valid. (As ≈explained in 22, another $(x, y)$ notation means you have to carry less in mind.)

I think you got the correct expression for $a$; when you have a in terms of $r$ then it is very immediate to get $b$.

 

[Ray Vickson]

Originally i denoted the width and height as a and b, see my drawing.
Later you,Ray, passed to denote width and height as 2a and 2b in post #15, since you wrote $~a = \sqrt{r^2-b^2}$

I returned, in post #21, to my original notation, i didn't make a mistake. please continue to help.

I didn't explain, in post #15, that after i found the correct answer as a=r, i continued to try to solve in StoneTemple's method GM<AM, post #5, and still cannot solve. that is what i explained in my last post. sorry for not explaining, i hurry to reply for you all not to loose contact and don't explain enough.

I have only time for skimming these posts, so I can be wrong. I'm fairly sure though that in #21 you have mixed two notations, "ours" and "yours" which is that of the original problem. I think the last equation In the first line of #21 is true in your notation, however then the → which annoyed Ray Is not valid. (As ≈explained in 22, another $(x, y)$ notation means you have to carry less in mind.)

I think you got the correct expression for $a$; when you have a in terms of $r$ then it is very immediate to get $b$.

OK: $(a/2)^2 +(b/2)^2 = r^2$ does give $b = \sqrt{4 r^2 - a^2}.$

Anyway, that is not the point: here is what YOU wrote in your first line of #23: $S = a b^3$, $a^2+b^2=r^3 \rightarrow S = a (4r^2-a^2)^{3/2}$. That is very much a mistake.

 

[Karol]

Anyway, that is not the point: here is what YOU wrote in your first line of #23: $S = a b^3$, $a^2+b^2=r^3 \rightarrow S = a (4r^2-a^2)^{3/2}$. That is very much a mistake.

You are right, now i see i didn't mean the right connection, very sorry again and thank you, Ray.

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

 

[epenguin]

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.

And in the final answer you need to bring back the k that was in #1

9^¼ is √3.

You seem to me to be making quite heavy weather of something elementary. Forget now about logs, of any kind, Just think you are looking at a schoolbook Introductory Algebra exercise 5 of chapter 2 ' substitutions'

Ex 5
i) If
$$S=a(4r^2-a^2)^{3/2}$$
and $a=r$

what is S in terms of $r$ ?

ii) ( from the other equations) what is $b$ in terms of $r$, and what is the relation between $a$ and $b$ ?

I do not see the point of your ≤ , we are looking for =. .