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

  • Thread starter PhysicsinCalifornia
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In summary: Thanks for your help. In summary, you need to find the critical points of the function and find the maximum point. After that, you'll need to solve for V'(r).One of the critical points is the maximum point, so you'll need to use the multiplication rule in the second term to differentiate the function.
  • #1
PhysicsinCalifornia
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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
[tex]V_s = \frac{4}{3}\pi r^3[/tex]
and the volume for the cone is
[tex]V_c = \frac{1}{3}\pi r^2 h[/tex]
(obviously)

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

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

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

Therefore
[tex] 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}[/tex]

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:
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  • #2
I guess you're right so far. You should try to calculate [tex]V'(r)[/tex] now. After that, you'll need to solve [tex]V'(r) = 0[/tex] 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.
 
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  • #3
tiagobt said:
I guess you're right so far. You should try to calculate [tex]V'(r)[/tex] now. After that, you'll need to solve [tex]V'(r) = 0[/tex] 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,

[tex]V_s = \frac{4}{3}\pi r^3[/tex]

[tex]V_c = \frac{1}{3}\pi r^2 h[/tex]

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"
 

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  • #4
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,

[tex]V_s = \frac{4}{3}\pi r^3[/tex]

[tex]V_c = \frac{1}{3}\pi r^2 h[/tex]

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 [tex]r[/tex] 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
 
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Related to Solving Max Volume Right Circular Cone Inscribed in Sphere Radius 3cm

1. How do you find the maximum volume of a right circular cone inscribed in a sphere with a radius of 3cm?

The maximum volume of a right circular cone inscribed in a sphere can be found by using the formula V = (1/3)πr^2h, where r is the radius of the base of the cone and h is the height of the cone. In this case, r = 3cm and h = 6cm (double the radius of the sphere). Plugging in these values, we get V = (1/3)π(3cm)^2(6cm) = 18π cm^3.

2. What is the relationship between the sphere and the cone in this problem?

The cone is inscribed in the sphere, meaning that the base of the cone touches the inner surface of the sphere. This creates a right angle between the base of the cone and the diameter of the sphere. The height of the cone is also equal to the radius of the sphere.

3. How does the volume of the cone change if the radius of the sphere is increased or decreased?

If the radius of the sphere is increased, the maximum volume of the cone will also increase. This is because a larger sphere will have a larger base for the cone, resulting in a larger volume. Conversely, if the radius of the sphere is decreased, the maximum volume of the cone will also decrease.

4. Can this formula be used for any size of sphere?

Yes, this formula can be used for any size of sphere as long as the cone is inscribed in the sphere. The radius of the sphere will affect the volume of the cone, but the formula will still hold true.

5. How can this problem be applied in real-life situations?

This problem can be applied in many real-life situations, such as in engineering and construction. For example, if a water tank needs to be designed to fit inside a cylindrical water tower, the maximum volume of the cone that can fit inside the tower can be calculated using this formula. This can help determine the size and capacity of the water tank needed.

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