What is the Maximum Volume of an Upside Down Cone Inside a Larger Cone?

[fubag]

Find the maximum volume of a right circular cone placed upside down in a right circular cone of radius R and height H. The volume of a cone of radius r and height h is 4/3pir^2h.

I need help starting this one up...

Any feedback will be appreciated.

 

[mjsd]

a picture or two may be needed for me to see the exact criteria as to how the cone must be placed inside the othe cone... or do u have to work that out during the optimization process too?

 

[fubag]

ok sorry

here is a crude picture i created...

the gray cone is the one upside down...

hope this helps.

http://img451.imageshack.us/img451/6751/optimizationfi3.th.jpg

 

[mjsd]

so the axis of symmetry must align?

shall need a function of h in terms of r or vice-versa where R, H are fixed variables.

 

[HallsofIvy]

Looked at "from the side" you have two triangles. You should be able to find the equation for the sides of the large triangle (i.e. large cone) and use that to find the radius of the base of the small cone as a function of its height.

 

[fubag]

ok so.. equation for the sides of the large triangle =

S = 2R + 2sqrt.(H^2 + R^2)

im not sure how to solve for the radius of the small cone as a functon of its height

 

[HallsofIvy]

How about drawing a picture? Where exactly does a cone of height h intersect that line- in other words, when h= H, what is R?

Your "equation for the sides of the large triangle makes no sense because it involves 3 variables. If "R" and "H" are the radius and height of the cone, what is "S"?

 

[fubag]

R is in respect to H, so R=r...?

 

[fubag]

ok

so i rethought about it

and drew a diagramand now I have

r = 1/2R

and h = 1/2H

so the next step is maximizing this...?

with respect to t, or x, or what?

V = 4/3pi (R/2)^2(H/2)
dV/dx?

how can I solve for the bigger R in terms of H?

 

[HallsofIvy]

No, r is not (1/2)R! Taking (0,0) as the point of the large cone, the axis of the cone as the y-axis, the right side of the large cone has equation y= (H/R)x so that when x= 0, y= 0, the point of the cone, and when x= R, y= H. Your small cone has its point at y= H. When x= r, its side touches that line. Of course, at that point, the y coordinate is H-h. Put x= r, y= H-h into the equation y= (H/R)x and you get H-h= (H/R)r or r= (R/H)(H-h)= R- (R/H)h.

You might find it simpler to write h as a function of r: h= (H/R)(R-r)= H- (H/R)r.

 

[fubag]

if i draw the larger cone as a triangle, and the smaller cone as an upside down triangle within the larger, cone, I can clearly see 4 triangle being formed...so in order to maximize the smaller cones area...I took 1/4 the area of the larger cone, using the formula A = 1/2bh...

that is how I got r = (1/2)R and h= (1/2)H.

but after looking at your last post, I got a little confused...even after acquiring the function of h in terms of r and H, how can I precede the question to find the maximum value of the smaller cone?

 

[HallsofIvy]

WHY did you take "1/4 the area of the larger cone"? The whole point of the question is to determine what the maximum area of the smaller cone can be. How do you know before you even start that it will be 1/4 of the area?

 

[fubag]

we are looking for maximum volume of the cone, not area...

but ok...I understand that I can't assume that...

I redrew the triangles, and this time saw why.my teacher hinted about finding the volume of the inner cone with respect to h..so now do I just use that equation of h(r) above and plug it back into the volume of a cone and take the derivative and set it equal to zero?

 

[HallsofIvy]

Excuse me. YOU were the one who said "so in order to maximize the smaller cones area...I took 1/4 the area of the larger cone, using the formula A = 1/2bh..."

 

[hotvette]

I think there is a typo on the volume of a cone - should be 1/3 instead of 4/3. Re how to solve, the most straightforward is:

1. Determine r or h in terms of H,R (simple trig w/ triangles)
2. Substitute into equation for V to get V = f(r) or V = f(h)
3. Take derivative of (2), set = 0, and solve

The result is a quadratic equation that is readily solved. The answer is surprising simple.