Find the cone transcribing a prism

1. Oct 30, 2013

Elpinetos

1. The problem statement, all variables and given/known data
Find the minimum value of the volume of a cone that is transcribing a four-sided prism with a=42cm and h=8cm

2. Relevant equations
$V=\frac{r^{2}H\pi}{3}$
$\frac{H}{r}=\frac{h}{r-\frac{a}{2}}$

3. The attempt at a solution
From the equation above it follows that
$H=\frac{2hr}{2r-a}$

Inserting this into V I get
$V=\frac{2hr^{3}\pi}{6r-3a}$

Taking the derivative of this I get
$\frac{dr}{dV}=\frac{2r^{2}\pi (4r-3a)}{(2r-a)^{2}}$

Setting this 0 and solving for r I get
$r=\frac{3a}{4}$

Since a is 42cm I get 31.5cm as the answer

Setting this into the H equation I get:
$H=3h$

Since h is 8cm, I get 24cm as the answer

Now, my solution textbook tells me that I got the answer for H right, but it tells me that r should be
$r=\frac{63\sqrt{2}}{2}$

What did I get wrong? Where does the square root come from? :/

2. Oct 30, 2013

haruspex

Pls define 'a'. Is it the length of each edge of the base of the prism (tetrahedron)?
How do you get the term r - a/2?

3. Oct 30, 2013

Elpinetos

Yeah, a is the length of the edges of the prism.
And I get the equation for H from the equivalence of the sides in similar triangles

4. Oct 30, 2013

LCKurtz

I don't understand this problem at all. If the prism is a regular tetrahedron with side 42, its height isn't 8. Does "transcribing" mean circumscribing? Is one face of the prism to be flat on the base of the cone? If so, don't see how it is an optimization problem in the first place.

Last edited: Oct 30, 2013
5. Oct 30, 2013

haruspex

It doesn't say regular tetrahedron, in fact it doesn't mention tetrahedra at all. Just not sure what is meant by a 'four sided prism'.
In my experience, prism usually refers to the projection into a third dimension of a polygon, but that implies at least five sides. Maybe it means the projection of a quadrilateral. That fits with having two different lengths, but then we'd also need to assume it's rectangular - in which case why doesn't it say rectangular?
So I presumed it to be a tetrahedron height h on a base which is an equilateral triangle of side a. But if so, you are right that the optimisation is obvious.
Elpinetos, none of those interpretations imply to me that r - a/2 is an interesting quantity. That's why I asked how you arrived at that, in the hope of understanding how you are interpreting the prism.

6. Oct 31, 2013

Elpinetos

I interpret it as a rectangle with a base with side lengths a and a height h.
Around that I'm putting a cone

Sorry, I'm translating those problems on-the-fly from another language, so I might not be completely acurate with my descriptions, seeing how I never studied math in english :D

From this image I get that H must behave to R as h does to R-a/2

7. Oct 31, 2013

haruspex

Ok, let's go with that interpretation. But your r-a/2 is wrong. Try drawing the view from above.

8. Oct 31, 2013

Elpinetos

Oh now I see, the blue line should actually be
$R-\frac{\sqrt{4R^{2}-a^{2}}}{2}$

Correct? :)

9. Oct 31, 2013

Elpinetos

I rethought the problem and see now that
$R=\frac{a\sqrt{2}}{2}$

Though I still don't see how that renders my
$\frac{H}{R}=\frac{h}{R-\frac{a}{2}}$
invalid.

I still see the same ratio :/

EDIT: It even says in the title of the page "Problems solvable with the intercept theorem"...

Last edited: Oct 31, 2013
10. Oct 31, 2013

haruspex

Your new diagram is wrong in a different way. At the base of the cone, where the radius is R, can it touch the rectangular block anywhere?

11. Oct 31, 2013

LCKurtz

At this point, I think the cross section in post #7 is correct and the work in the original post is correct. The radius comes out 63/2 and the answer key has an incorrect square root in the denominator.

12. Oct 31, 2013

haruspex

I don't see how. It shows a width of a for the rectangular block. The cone must touch the upper corners of the block, and opposite pairs of those are distance a√2 apart.

13. Oct 31, 2013

LCKurtz

Ahhh. You're right; that is a bit tricky to visualize. The block needs to be rotated $45^\circ$ about its vertical axis showing 3 vertical edges for the cross section to be correct.

14. Nov 1, 2013

LCKurtz

In case anyone is still watching, here are a couple views (not to scale) of what this looks like. The first shows a front view looking directly at a face of the prism.

This is what the top view looks like for the same orientation.

15. Nov 5, 2013

Elpinetos

Hey, sorry, I've been busy the last couple of days
The figure you posted pretty much corresponds to my figure in post #6, where I still fail to see why H/R = h/R-a/2 is incorrect
What would be the correct side condition?

16. Nov 5, 2013

LCKurtz

No it doesn't. The top view in my post #14 looks like your figure in post #9.

Try to get that from the front view (first figure) in post #14. Your similar triangles aren't there.

17. Nov 5, 2013

haruspex

Have you read post #12? Do you understand what I wrote?

18. Nov 7, 2013

Elpinetos

Ahhhhhh now it makes sense.
What I calculated was basically for a pyramid, not a cone, wasn't it? :)

Though, where is the similar triangle stated in the overview of the problems?
The problem is listed under "solve these using similar triangle relationships"

19. Nov 7, 2013

haruspex

Take a vertical slice that passes through two diagonally opposite corners of the cube.