Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.

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Understand the Problem

The question is asking us to find the volume of the largest right circular cone that can be inscribed in a sphere with a radius of 3. To solve this, we need to use the formula for the volume of a cone, V = (1/3)πr²h, and relate the dimensions of the cone to the radius of the sphere, taking into account the constraints provided by the inscribed shape.

Answer

The maximum volume of the cone is $9\pi$.
Answer for screen readers

The volume of the largest right circular cone that can be inscribed in a sphere of radius 3 is $9\pi$.

Steps to Solve

  1. Identify the dimensions of the cone

    Let the height of the cone be $h$ and the radius of the circular base be $r$. According to the problem, the sphere has a radius of 3. We can relate the height $h$ and the base radius $r$ using the Pythagorean theorem.

  2. Relate the cone's dimensions to the sphere's radius

    In the diagram, we have a vertical height $y$ (from the apex of the cone to the center of the sphere) and a distance $x$ (from the center of the sphere to the bottom of the cone). The total height relates as follows: [ h = y + x ] Using the relationship from the Pythagorean theorem for a right triangle formed: [ 3^2 = r^2 + y^2 \quad \text{which simplifies to} \quad 9 = r^2 + y^2 ]

  3. Express volume of the cone

    The volume $V$ of the cone can be expressed using the formula for the volume of a cone: [ V = \frac{1}{3} \pi r^2 h ] We can substitute $h$ in terms of $y$ and $x$: [ V = \frac{1}{3} \pi r^2 (y + x) ]

  4. Express $x$ in terms of $y$

    From the previous step, since $h = y + x$ and $y = 3 - x$, we can express $x$ in terms of $y$: [ x = 3 - y ]

  5. Substitute and simplify the volume function

    Substitute the expressions for $h$ and $r$ into the volume equation: [ V = \frac{1}{3} \pi r^2 (y + (3 - y)) = \frac{1}{3} \pi r^2 \cdot 3 = \pi r^2 ] From $9 = r^2 + y^2$, we know $r^2 = 9 - y^2$. Substituting this into the volume, we can express $V$ only in terms of $y$: [ V(y) = \pi (9 - y^2) ]

  6. Optimize the volume

    To find the maximum volume, we need to differentiate $V(y)$ with respect to $y$ and set the derivative equal to zero: [ V'(y) = \pi (-2y) = 0 \quad \Rightarrow \quad y = 0 ] This yields endpoints to check: ( y = 0 ) and ( y = 3 ).

  7. Calculate the volume

    We will use the maximum found: At ( y = 3 ): [ V = \pi (9 - 3^2) = \pi (9 - 9) = 0 ] At ( y = 0 ): [ V = \pi (9 - 0) = 9\pi ]

The volume of the largest right circular cone that can be inscribed in a sphere of radius 3 is $9\pi$.

More Information

The cone's maximum volume occurs at the height of the cone reaching the sphere's equator where the radius is maximized. The calculation is based on geometric relationships derived from the cone’s inscribed position within the sphere.

Tips

  • Confusing the dimensions of the cone and incorrectly applying the Pythagorean theorem can lead to wrong relationships between height and radius.
  • Not checking endpoints during optimization may miss the maximum volume possible.
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