Finding the ratio of the volumes of a cube and a sphere

[Karol]

Homework Statement

Homework Equations

Volue of a sphere: $~\displaystyle V=\frac{4}{3}\pi r^3$
Area of a sphere: $~\displaystyle A=4\pi r^2$
Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution

The fixed area is k, the edge is a:
$$6a+4\pi r^2=k~\rightarrow~a=\frac{k-4\pi r^2}{6}$$
$$V=a^3+\frac{4}{4}\pi r^3=\left[ \frac{k-4\pi r^2}{6} \right]^3+\frac{4}{4}\pi r^3$$
$$V'=\frac{3}{216}(k-4\pi r^2)^2(-8)\pi r+4\pi r^2$$
$$V'=0:~~4\pi r-\frac{8}{72}(k-4\pi r^2)^2\pi=0~\rightarrow~r=\frac{3k-2}{12\pi}$$
$$\frac{a}{r}=\frac{(k-4\pi r^2)12\pi}{6(3k-2)}=\frac{\pi(k-4\pi r^2}{3k-2}$$
The k doesn't cancel

 

[Orodruin]

The fixed area is k, the edge is a:
$$6a+4\pi r^2=k~\rightarrow~a=\frac{k-4\pi r^2}{6}$$

This is dimensionally inconsistent.

To be honest, many of your errors in all of your similar threads could be avoided if you:

• Were more careful in your arithmetic.
• Were more careful in the construction of your basic expressions.
• Checked your expressions for dimensional consistency.

 

[PeterDonis]

Moderator's note: Thread title adjusted to be more descriptive and less generic.

@Karol, please avoid choosing thread titles that are (a) too generic, and (b) too similar to the titles of previous threads.

 

[PeterDonis]

The k doesn't cancel

That's because your formulas for both total area and total volume are wrong. As @Orodruin says, you need to check your work more carefully.

 

[Ray Vickson]

Homework Statement

View attachment 222596

Homework Equations

Volue of a sphere: $~\displaystyle V=\frac{4}{3}\pi r^3$
Area of a sphere: $~\displaystyle A=4\pi r^2$
Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution

The fixed area is k, the edge is a:
$$6a+4\pi r^2=k~\rightarrow~a=\frac{k-4\pi r^2}{6}$$
$$V=a^3+\frac{4}{4}\pi r^3=\left[ \frac{k-4\pi r^2}{6} \right]^3+\frac{4}{4}\pi r^3$$
$$V'=\frac{3}{216}(k-4\pi r^2)^2(-8)\pi r+4\pi r^2$$
$$V'=0:~~4\pi r-\frac{8}{72}(k-4\pi r^2)^2\pi=0~\rightarrow~r=\frac{3k-2}{12\pi}$$
$$\frac{a}{r}=\frac{(k-4\pi r^2)12\pi}{6(3k-2)}=\frac{\pi(k-4\pi r^2}{3k-2}$$
The k doesn't cancel

Since when has the area of a cube of side-length $a$ become $6 a$? Shouldn't the $a$ be squared?

Or, you could let the area of one face be $a$, but then the side would be $\sqrt{a}$.

I have told you many times before in numerous other threads: you need to be more careful. You will NEVER succeed if you persist in continually making such elementary errors.

 

[Orodruin]

So Ray put it more straight up than I did, but the idea is the same. You keep making errors that you should easily be able to catch if you checked your work. This makes it appear as if you have speeded through the problem rather sloppily, predictably made a mistake that you do not want to spend time on finding, and posted it all here for us to troubleshoot. An important part of learning all of this is checking your own work and finding mistakes. By not doing so you are just hurting your own learning and wearing out helpers so that they become less likely to help you once you really run into trouble.

 

[Karol]

Correct. i will try to check before i post.
$$6a^2+4\pi r^2=k~\rightarrow~a=\sqrt{\frac{k-4\pi r^2}{6}}$$
$$V=a^3+\frac{4}{3}\pi r^3=\left[ \frac{k-4\pi r^2}{6} \right]^{3/2}+\frac{4}{3}\pi r^3$$
$$V'=\frac{3}{2}\left[ \frac{k}{6}-\frac{2\pi}{3}r^2 \right]^{1/2}\left( -\frac{4\pi}{3} \right)r+4\pi r^2$$
$$V'=0:~~r=\frac{1}{2}\left[ \frac{k}{6}-\frac{2\pi}{3}r^2 \right]^{1/2}~\rightarrow~r=\sqrt{\frac{k}{6+\pi}}$$
The edge a contains r so the ratio r/a isn't correct

 

[Orodruin]

The edge a contains r so the ratio r/a isn't correct

Again, you have not thought this through.

 

[Karol]

$$a=\sqrt{\frac{k-4\pi r^2}{6}}=\sqrt{\frac{k-4\pi\frac{k}{6+\pi}}{6}}=\sqrt{\frac{(2-\pi)k}{2(6+\pi)}}$$
$$\frac{r}{a}=\sqrt{\frac{2-\pi}{2}}$$
Wrong, and i can't distinguish between maximum and minimum.
The answers:

 

[Orodruin]

Because somewhere you have made an arithmetic error. Just plotting your volume function for some k, you could conclude that the minimum is not where you found it.

 

[Ray Vickson]

$$a=\sqrt{\frac{k-4\pi r^2}{6}}=\sqrt{\frac{k-4\pi\frac{k}{6+\pi}}{6}}=\sqrt{\frac{(2-\pi)k}{2(6+\pi)}}$$
$$\frac{r}{a}=\sqrt{\frac{2-\pi}{2}}$$
Wrong, and i can't distinguish between maximum and minimum.
The answers:
View attachment 222610

You can use a second-derivative test to distinguish between a (local) max vs. min. However, you should definitely avoid trying that until you have solved the first-order problem correctly.

Also: you need worry about the fact that your $r$ is restricted to a finite interval, so the derivative might not vanish at a global max or min.

 

[PeterDonis]

Regarding the first-order problem, I'm not sure that finding expressions as a function of $r$ and taking their derivatives is actually what is needed. The real question is where the maximum and minimum of the total volume are as a function of the ratio of the edge of the cube to the diameter of the sphere. If we call that ratio $\rho$, then it seems to me that what is needed is an expression for the total volume as a function of $\rho$.

 

[Orodruin]

Also, based on how many of this type of problems you have been looking at, you should consider learning the method of Lagrange multipliers.

 

[Ray Vickson]

Regarding the first-order problem, I'm not sure that finding expressions as a function of $r$ and taking their derivatives is actually what is needed. The real question is where the maximum and minimum of the total volume are as a function of the ratio of the edge of the cube to the diameter of the sphere. If we call that ratio $\rho$, then it seems to me that what is needed is an expression for the total volume as a function of $\rho$.

That will not work; somehow the constant value of total area needs to come into the expressio.

 

[PeterDonis]

somehow the constant value of total area needs to come into the expressio

It does. You start out with two quantities that could potentially vary, the ratio $\rho$ and the diameter $D$ of the sphere. You use the area equation to eliminate $D$ by finding an expression for it in terms of $\rho$ and the constant total area; that basically eliminates one degree of freedom in the problem by imposing a constraint. Then you can derive an equation for the total volume as a function of $\rho$ and the total area; $\rho$ is then the only variable.

 

[Orodruin]

It does. You start out with two quantities that could potentially vary, the ratio $\rho$ and the diameter $D$ of the sphere. You use the area equation to eliminate $D$ by finding an expression for it in terms of $\rho$ and the constant total area; that basically eliminates one degree of freedom in the problem by imposing a constraint. Then you can derive an equation for the total volume as a function of $\rho$ and the total area; $\rho$ is then the only variable.

But this is just a change of variables relative to what the OP has attempted.

 

[PeterDonis]

this is just a change of variables relative to what the OP has attempted

Yes, agreed. I'm just wondering if it might be easier to "read off" the behavior if everything is expressed as a function of $\rho$.

 

[LCKurtz]

Does anyone except me find it surprising that the minimum occurs when the cube edge equals the diameter of the sphere?

 

[Karol]

$$V'=\frac{3}{2}\left[ \frac{k}{6}-\frac{2\pi}{3}r^2 \right]^{1/2}\left( -\frac{4\pi}{3} \right)r+4\pi r^2=4\pi\left[ r^2-\frac{1}{2}\left( \frac{k-4\pi r^2}{6}\right)^{1/2}\right]$$
$$V'=0:~~r=\frac{1}{2}\left[ \frac{k}{6}-\frac{2\pi}{3}r^2 \right]^{1/2}~\rightarrow~r=\sqrt{\frac{k}{4(6+\pi)}}$$
$$2r=\sqrt{\frac{4k}{4(6+\pi)}}=\sqrt{\frac{k}{6+\pi}}$$
$$a=\sqrt{\frac{k-4\pi r^2}{6}}=\sqrt{\frac{k-4\pi\frac{k}{4(6+\pi)}}{6}}=\sqrt{\frac{k}{6+\pi}}~\rightarrow~\frac{a}{2r}=1$$
The second derivative:
$$V''=4\pi \left[ 2r-\frac{1}{2} \left( \frac{k-4\pi r^2}{6}\right)^{-1/2} \left( -\frac{8\pi}{6} \right) r \right]=4\pi r\left[ 2-\frac{\pi}{3}\left( \frac{k-4\pi r^2}{6}\right)^{-1/2} \right]$$
$$V''\left( \sqrt{\frac{k}{4(6+\pi)}} \right)=4\pi \sqrt{ \frac{k}{4(6+\pi)}} \left[ 2-\frac{\pi}{3}\left( \frac{k-4\pi \frac{k}{4(6+\pi)}}{6} \right)^{-1/2} \right]$$
$$V''=2\pi \sqrt{\frac{k}{4(6+\pi)}} \left[ 2-\frac{\pi}{3} \sqrt{\frac{6+\pi}{k}} \right]$$
The last parentheses depend on k, so how know if V'' is negative or positive?
The $~\rho=\frac{a}{2r}~$ method:
$$\rho=\frac{a}{2r}~\rightarrow~r^2=\frac{a^2}{4\rho ^2},~~a=\sqrt{\frac{k-4\pi r^2}{6}}$$
$$a=\sqrt{\frac{k-4\pi \frac{a^2}{4\rho ^2}}{6}}~\rightarrow~a=\rho\sqrt{\frac{k}{6\rho^2+\pi}}$$
$$r^2=\frac{a^2}{4\rho^2}=\frac{k}{4(6\rho^2+\pi)}$$
$$V=a^3+\frac{4}{3}\pi r^3=\rho^3\left( \frac{k}{4(6\rho^2+\pi} \right)^{3/2}+\frac{4\pi}{3}\left( \frac{k}{4(6\rho^2+\pi} \right)^{3/2}$$
Complicated

 

[Orodruin]

Again you end up with something dimensionally inconsistent. For the last time, check your work for obvious errors!

 

[StoneTemplePython]

my take would be to look for a (quasi) linear representation of the problem. You lose some units interpretability this way so I'm not going to post much right now and I'd suggest Karol complete existing methods.

Hopefully the problem is solved in coming days and I can post over the weekend.
- - - -
A sketch:

Geometrically the approach is quite nice (and I'm wondering if OP is not using pictures enough?). I think this approach is a lot easier to interpret if you work through graphically

The idea is define:

$s_1 := 6a^2$
$s_2 := 4\pi r^2$

hence the surface area constraint is linearized:

$s_1 + s_2 = c$

note that $s_1 \geq 0$ and $s_2 \geq 0$ and the problem statement indicates this applies for any $c \gt 0$ so it may be instructive to consider setting $c := 1$ and also consider $c := \delta$, for some small $\delta \gt 0$, though these special cases aren't needed per se.

$\text{Total Volume Sum} = f( s_1 ) +g(s_2)$

$f$ and $g$ are both differentiable, and strictly convex over positive numbers.

The minimization then follows from either (a) interpretting the respective tangent lines special role as being the best (linear) approximation of our functions $f$ and $g$ over some small neighborhood and/or (b) interpretting the tangent lines as the linear lower bounds of convex functions $f$ and $g$. The argument via (a) is a bit looser but probably more intuitive, while the argument via (b) is strict.

The maximization is even easier.

 

[Orodruin]

As stated in #13, let me make another pitch for the method of Lagrange multipliers, which solves this problem in about three lines (plus the check of which of the bounding limits is the other extreme) without any need for differentiating square roots.

 

[Karol]

$\rho~$ is dimensionless, k[m²] so the units of:
$$\rho^3\left( \frac{k}{4(6\rho^2+\pi} \right)^{3/2}=[m^2]^{3/2}=[m^3]$$
Are of volume

 

[Orodruin]

$\rho~$ is dimensionless, k[m²] so the units of:
$$\rho^3\left( \frac{k}{4(6\rho^2+\pi} \right)^{3/2}=[m^2]^{3/2}=[m^3]$$
Are of volume

I am not talking about the part with $\rho$. Doing that $\rho$ will always be dimensionless and you will lose the possibility to do dimensionsl analysis to check your answers.

 

[Karol]

I had a mistake in V':
$$V'=\frac{3}{2}\left[ \frac{k}{6}-\frac{2\pi}{3}r^2 \right]^{1/2}\left( -\frac{4\pi}{3} \right)r+4\pi r^2=4\pi r \left[ r-\frac{1}{2}\left( \frac{k-4\pi r^2}{6}\right)^{1/2}\right]$$
$$V'' = 4\pi\left[ r - \frac{1}{2} \sqrt{ \frac{ k-4 \pi r^2}{6} }+r\left( 1-\frac{\pi r}{3\sqrt{\frac{k-4\pi r^2}{6}}} \right) \right]$$
Now the dimensions of all the parenthesis of V'' are [m], correct.
$$V''\left( r=\frac{1}{2}\sqrt{\frac{k}{6+\pi}} \right)=\frac{\pi(6-\pi)}{3}\sqrt{\frac{k}{6+\pi}}>0$$
So it's a minimum.
$$V(a=0)=4\pi r^2=\frac{\pi}{6+\pi}k$$
$$V(r=0)=a^3=\left( \frac{k}{6} \right)^{3/2}$$
How to know which is bigger? they depend on k.
I don't understand:

The minimization then follows from either (a) interpreting the respective tangent lines special role as being the best (linear) approximation of our functions fff and ggg over some small neighborhood and/or (b) interpreting the tangent lines as the linear lower bounds of convex functions $f$ and $g$. The argument via (a) is a bit looser but probably more intuitive, while the argument via (b) is strict.

 

[Orodruin]

$$V(a=0)=4\pi r^2=\frac{\pi}{6+\pi}k$$
$$V(r=0)=a^3=\left( \frac{k}{6} \right)^{3/2}$$
How to know which is bigger? they depend on k.
I don't understand:

Again a trivial mistake. Double check your results.

 

[Karol]

$$V(r=0)=a^3=\left( \frac{k}{6} \right)^{3/2}$$
$$a=0~\rightarrow~4\pi r^2=k~\rightarrow~r^2=\frac{k}{4\pi}$$
$$V(a=0)=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\left( \frac{k}{4\pi} \right)^{3/2}$$
On the a=0 side, $~\frac{4\pi}{3}>1~$ but $~\frac{k}{4\pi}<\frac{k}{6}$
So how to know which is bigger?

 

[Orodruin]

Just compare the numbers?

 

[Karol]

$$V(r=0)=a^3=\left( \frac{k}{6} \right)^{3/2}=0.07\sqrt{k^3}$$
$$V(a=0)=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\left( \frac{k}{4\pi} \right)^{3/2}=0.09\sqrt{k^3}$$
The maximum V is at ratio:
$$V(a=0)>V(r=0)~\rightarrow~\frac{a}{2r}=0$$

$s_1 := 6a^2$
$s_2 := 4\pi r^2$
hence the surface area constraint is linearized:
$s_1 + s_2 = c$
note that $s_1 \geq 0$ and $s_2 \geq 0$ and the problem statement indicates this applies for any $c \gt 0$ so it may be instructive to consider setting $c := 1$ and also consider $c := \delta$, for some small $\delta \gt 0$, though these special cases aren't needed per se.
$\text{Total Volume Sum} = f( s_1 ) +g(s_2)$
$f$ and $g$ are both differentiable, and strictly convex over positive numbers.
The minimization then follows from either (a) interpretting the respective tangent lines special role as being the best (linear) approximation of our functions $f$ and $g$ over some small neighborhood and/or (b) interpretting the tangent lines as the linear lower bounds of convex functions $f$ and $g$. The argument via (a) is a bit looser but probably more intuitive, while the argument via (b) is strict.
The maximization is even easier.

$$V=\frac{r}{3}(s_2)+6a(s_1)$$
Now what?

 

[StoneTemplePython]

$$V=\frac{r}{3}(s_2)+6a(s_1)$$
Now what?

surely what you meant to write instead is:

$$V=\frac{r}{3}(s_2)+\frac{1}{6}a(s_1)$$- - - - -
but in order to make use of units and strict convexity what I actually have in mind is

$f(x) = \big(\frac{x}{6}\big)^\frac{3}{2} = \frac{x^\frac{3}{2}}{6^\frac{3}{2}}$
and
$g(x) = \Big(\frac{x^\frac{3}{2}}{6 \pi^\frac{1}{2}}\Big)$

(confirm that these are correct and give the right volume equations for $f(s_1) + g(s_2)$ )

and remember the problem is that $s_1 + s_2 = c$ where $c\gt 0$ and $s_1, s_2 \geq 0$
- - - -
To start easy:

For the maximization problem:

First: show that for any $x\gt 0$

$f(x) \lt g(x)$

and confirm that $g(0) = 0 = f(0)$

Second

Thus, for the maximization problem, for real non-negative numbers -- call them $s_1$, $s_2$, we have the below inequality

$f(s_1) + g(s_2) \leq g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c) = f(0) + g(c)$

The left inequality is immediate because
$f(s_1) \leq g(s_1)$
and we then add $g(s_2)$ to each side to get $f(s_1) + g(s_2) \leq g(s_1) + g(s_2)$

The second inequality needs proven: why is it that
$g(s_1) + g(s_2) \leq g(s_1 + s_2)$

After you have that proven, the maximization problem is done -- i.e. the Right Hand Side of the inequality tells you that you get the best possible 'score' when you allocate everything to $g$ which maps to sphere volume.