Solving Max Volume Right Circular Cone Inscribed in Sphere Radius 3cm

  • Thread starter Thread starter PhysicsinCalifornia
  • Start date Start date
AI Thread Summary
The discussion revolves around solving an optimization problem involving a right circular cone inscribed in a sphere with a radius of 3 cm. Participants focus on finding the cone's dimensions and volume using calculus, specifically by differentiating the volume function with respect to the cone's base radius. There is clarification on the correct terminology for mathematical operations, emphasizing the difference between "derive" and "differentiate." The conversation also addresses potential confusion caused by using the same variable for different values in the equations. Overall, the thread highlights the collaborative effort to solve the problem and correct misunderstandings.
PhysicsinCalifornia
Messages
58
Reaction score
0
HW problem--NEED HELP!

This is an optimization problem with differential calc. I really need help with this

Here's the prob:

A right circular cone is being inscribed in a sphere of radius 3cm.

Find
a) the dimensions (base radius, and height) of the right circular cone with the largest volume
b) the cone's volume

Here's what I got::

Volume for sphere is
V_s = \frac{4}{3}\pi r^3
and the volume for the cone is
V_c = \frac{1}{3}\pi r^2 h
(obviously)

Now i got that h = 3+x
*Note that I cannot draw a pic, so it's hard to describe

Also, x = \sqrt{9 - r^2} using pythagorean's theorem

so the height would equal the radius of the sphere, 3 cm, plus x

Therefore
V(r) = \frac{1}{3} \pi r^2 (3 + \sqrt{9- r^2}) = \pi r^2 + \frac{\pi r^2 \sqrt{9 - r^2}}{3}

leaving everything in terms of r because we want the largest volume

How do I work it from here?

Thanks for your help in advance
 
Last edited:
Physics news on Phys.org
I guess you're right so far. You should try to calculate V'(r) now. After that, you'll need to solve V'(r) = 0 in order to find the critical points of the function. One of them should be the maximum point. In order to differentiate the function, you'll need to use the multiplication rule in the second term.
 
Last edited:
tiagobt said:
I guess you're right so far. You should try to calculate V'(r) now. After that, you'll need to solve V'(r) = 0 in order to find the critical points of the function. One of them should be the maximum point. To derive the function, you'll need to use the multiplication rule in the second term.

Wow! How did this problem go undetected for a couple of days? Good for you for digging back far enough to find it. I was at first bothered by the use of r twice to mean different things in the original equations,

V_s = \frac{4}{3}\pi r^3

V_c = \frac{1}{3}\pi r^2 h

but since the first one is a constant r = 3 and is used correctly in the formulation of the volume of the cone, it is not a problem. I drew a diagram for myself, so I'll pass it along.

An observation about your reply- The verb "derive" in your statement "To derive the function, you'll need to use the multiplication rule in the second term" should be "differentiate". Derive does not mean to take the derivative of. In mathematics it means "To arrive at by reasoning; deduce or infer: derive a conclusion from facts"
 

Attachments

  • cone.jpg
    cone.jpg
    18.5 KB · Views: 461
OlderDan said:
Wow! How did this problem go undetected for a couple of days? Good for you for digging back far enough to find it. I was at first bothered by the use of r twice to mean different things in the original equations,

V_s = \frac{4}{3}\pi r^3

V_c = \frac{1}{3}\pi r^2 h

but since the first one is a constant r = 3 and is used correctly in the formulation of the volume of the cone, it is not a problem. I drew a diagram for myself, so I'll pass it along.

An observation about your reply- The verb "derive" in your statement "To derive the function, you'll need to use the multiplication rule in the second term" should be "differentiate". Derive does not mean to take the derivative of. In mathematics it means "To arrive at by reasoning; deduce or infer: derive a conclusion from facts"
Yes, at first I was bothered with the use of the letter r for two different things too. But PhysicsinCalifornia didn't mess things up. I'm sorry for the misuse of the word "derive". English is not my mother language and sometimes I say things that sound weird... I'm going to edit my reply though.

Tiago
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top