# Thread: Regular hexagonal pyramid-maximum volume?

1. ## Regular hexagonal pyramid-maximum volume?

Guys has six bars in length of 7 meters from which he wants to build a teepee(tipis) in the shape of a regular hexagonal pyramid. Determine the height h (a base b) so that the teepee has as much space as possible.(By space I think they in a real mean volume....)
Any idea how to get the max.volume? Using the derivation operation? Something else? :-)

2. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by karelkop
Guys has six bars in length of 7 meters from which he wants to build a teepee(tipis) in the shape of a regular hexagonal pyramid. Determine the height h (a base b) so that the teepee has as much space as possible.(By space I think they in a real mean volume....)
Any idea how to get the max.volume? Using the derivation operation? Something else? :-)

$V=\dfrac{\sqrt 3} 2 s^2 h$

where $s$ is the length of a side, and $h$ is the altitude.

To construct this pyramid you need 6 pieces of bar length $s$, and 6 pieces of bar length $\sqrt{s^2 + h^2}$

You don't want to waste any bar material. By symmetry of the problem you are best off splitting each bar into two pieces, one will be a side, and one will be the slanted vertical piece.

So you have a constraint equation

$s+\sqrt{s^2+h^2}=7$

Now you can either solve for say h in terms of s and substitute that back into the Volume formula and then minimize that in the usual way by equating the first derivative to zero etc. Or, you can use the method of Lagrange multipliers.

Can you take it from here?

3. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by karelkop
Guys has six bars in length of 7 meters from which he wants to build a teepee(tipis) in the shape of a regular hexagonal pyramid. Determine the height h (a base b) so that the teepee has as much space as possible.(By space I think they in a real mean volume....)
Any idea how to get the max.volume? Using the derivation operation? Something else? :-)
Hello,

I've made a sketch of the situation. I'll refer to my labeling:

1. $\displaystyle R = s$ (property of a regular hexagon)
2. $\displaystyle V = \frac13 \cdot A \cdot H$ (A: base area; H: height of the pyramid)
3. $\displaystyle A = 6\cdot \frac12 \cdot R \cdot R \cdot \sin(60^\circ) = \frac32 \cdot R^2 \cdot \sqrt{3}$
4. $\displaystyle H^2 + R^2=e^2~\implies~R^2 = e^2-H^2$

Therefore you'll get V as a funktion in H:

$\displaystyle V(H)= \frac32 \cdot (e^2-H^2) \cdot \sqrt{3} \cdot H = \frac32 \cdot \sqrt{3} \cdot e^2 \cdot H - \frac32 \cdot \sqrt{3} \cdot H^3$

Replace e by 7.

Differentiate V(H) wrt H and solve for H the equation V'(H)=0

$\displaystyle \frac32 \cdot \sqrt{3} \cdot 49-\frac92 \cdot \sqrt{3} \cdot H^2 = 0~\implies~H=\frac73 \cdot \sqrt{3}\approx 4.04$

Now determine the maxiumum of the volume.

4. ## Re: Regular hexagonal pyramid-maximum volume?

I think you've ignored the material needed for the sides of the base.

5. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by romsek
I think you've ignored the material needed for the sides of the base.
Hello,

well, honestly, I've some difficulties to understand this
Determine the height h (a base b) so that ...
I noticed your solution so I added my calculations only in case the poor Sioux wants to sleep on the bare ground without any surrounding bars.

6. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by earboth
Hello,

well, honestly, I've some difficulties to understand this

I noticed your solution so I added my calculations only in case the poor Sioux wants to sleep on the bare ground without any surrounding bars.
actually I think your solution is probably correct since they don't mention cutting the bars at all. Trust an engineer to overthink the problem!

7. ## Re: Regular hexagonal pyramid-maximum volume?

Gents,just busy as a babysitter,but I will check your answers in the evening.
Just quickly - no cutting,the target is when you have one dimension (fixed) -you can change height or base, doesn ' t matter they are depending on each other-but when you change these,volume will change-the target is to get answer- what is the dimension of a base and height when I want to get maximum volume....
Thank you very much for now!

8. ## Re: Regular hexagonal pyramid-maximum volume?

Hi Earboth, I do really appreciate your response. I think I do understand what you saying (at least partly):-), but I still do have a question:
your formula:V = 1/3.A.H (A: base area; H: height of the pyramid) is clear for me....(it also exist V=H.a.R which is more nice to me but letīs work with yours)

for me A=3Ra((6.(s.a/2)).....s=R

anyway if I will calculate with your formulas,please check my attached file...didnīt you miss 1/3 in your formula for V?

9. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by karelkop
Hi Earboth, I do really appreciate your response. I think I do understand what you saying (at least partly):-), but I still do have a question:
your formula:V = 1/3.A.H (A: base area; H: height of the pyramid) is clear for me....(it also exist V=H.a.R which is more nice to me but letīs work with yours)

for me A=3Ra((6.(s.a/2)).....s=R

anyway if I will calculate with your formulas,please check my attached file...didnīt you miss 1/3 in your formula for V?
Hello,

thanks for the flowers!

Yes, you are right, there is indeed the factor $\displaystyle \frac13$ missing.

Corrected version:

$\displaystyle V(H)= \frac13 \cdot \frac32 \cdot (e^2-H^2) \cdot \sqrt{3} \cdot H = \frac12 \cdot \sqrt{3} \cdot e^2 \cdot H - \frac12 \cdot \sqrt{3} \cdot H^3$

Differentiate and solve for H:

$\displaystyle \frac12 \cdot \sqrt{3} \cdot 49-\frac32 \cdot \sqrt{3} \cdot H^2 = 0~\implies~H=\frac73 \cdot \sqrt{3}\approx 4.04$

That means my previous result was OK.

Plug in this value into the equation of the volume:

$\displaystyle V(H)= \frac12 \cdot (e^2-H^2) \cdot \sqrt{3} \cdot H \\ ~\implies~V(\frac73 \cdot \sqrt{3})= \frac12 \cdot (7^2-(\frac73 \cdot \sqrt{3})^2) \cdot \sqrt{3} \cdot (\frac73 \cdot \sqrt{3})=\frac{686}9\ m^3 \approx 76.22\ m^3$

This is the maximum volume you can construct with these bars, 7 m long.

10. ## Re: Regular hexagonal pyramid-maximum volume?

On the other hand I was calculating different way but now I am checking my notes I made yesterday night and I got the same result :-) so even if you missed something or not, I got the same result.
see pics attached...anyway I donīt understand to yours :"Now determine the maxiumum of the volume"

11. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by karelkop
On the other hand I was calculating different way but now I am checking my notes I made yesterday night and I got the same result :-) so even if you missed something or not, I got the same result.
see pics attached...anyway I donīt understand to yours :"Now determine the maxiumum of the volume"
Hello,

you wrote:
Determine the height h (a base b) so that the teepee has as much space as possible.(By space I think they in a real mean volume....)
Any idea how to get the max.volume? Using the derivation operation? Something else? :-)
Your calculations give you the length of the height which will produce the maximum volume, but you don't know the value of the volume. You must use the value of the height to determine the value of the volume. That's what I've asked you to do.

EDIT: I'm offline now: Good night to you!

12. ## Re: Regular hexagonal pyramid-maximum volume?

Oh now it is clear to me :-)
anyway althought I got the same value as you for H, my R=5.72 and a=4,93 so total value is 114,1 ... :-(

13. ## Re: Regular hexagonal pyramid-maximum volume?

I found it, you have small mistake in your last counting there should be 686/6 nad it is 114.....

man wishing you a good night and sending my many thanks !!!!!

14. ## Re: Regular hexagonal pyramid-maximum volume?

Originally Posted by karelkop
I found it, you have small mistake in your last counting there should be 686/6 nad it is 114.....

man wishing you a good night and sending my many thanks !!!!!
Good morning,

good job!

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