Question

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 13. v=cubic units (Round to two decimal places needed. Tutoring Help me solve this Get more help Clear al

Answer 1

To find the volume of the largest right circular cone inscribed in a sphere, we need to use the formula V = (2/3)πR³, where R is the radius of the sphere. Substituting the value of R = 13, the volume is approximately 1434.33 cubic units.

Step-by-step explanation:

To find the volume of the largest right circular cone inscribed in a sphere, we need to first understand the relationship between the cone and the sphere. The largest cone that can be inscribed in a sphere will have its height equal to the diameter of the sphere and will be tangent to the sphere at its apex.

Let's denote the radius of the sphere as R. The height of the cone (h) will be equal to 2R, and the radius of the cone (r) will be equal to R. Now we can find the volume of the cone using the formula:

V = (1/3)πr²h

Substituting the values, we get:

V = (1/3)π(R²)(2R)

V = (2/3)πR³

Now, substitute the value of the radius of the sphere (R = 13) to find the volume:

V = (2/3)π(13)³

V ≈ 1434.33 cubic units

Answer 2

The volume of the largest cone inscribed in a sphere of radius 13 is calculated by first using the Pythagorean theorem to relate the cone's height and radius, then applying calculus to find the radius that maximizes the cone's volume. The volume formula V = (1/3)πr2h is then used with the optimal radius to find the maximum volume, rounded to two decimal places.

Step-by-step explanation:

To find the volume of the largest right circular cone inscribed in a sphere of radius 13, we must first recognize that the center of the sphere will also be the vertex of the cone and the cone's base will be in contact with the interior surface of the sphere.

Let's denote the height of the cone as 'h' and the radius of the cone's base as 'r'. Since the cone is right circular, we can create a right triangle by slicing the cone and sphere in half, along the cone's axis of symmetry. In this triangle, the hypotenuse is the sphere's radius (13 units), and the two legs are the cone's height (h) and radius (r).

Using the Pythagorean theorem, we have:

h2 + r2 = 132

To maximize the volume of the cone, we need to take the derivative of the volume with respect to 'r' and set it to zero to find the critical points. The volume of a cone is given by V = (1/3)πr2h. Since h can be expressed in terms of r, we have V = (1/3)πr2(√(132 - r2)).

After finding the value of 'r' that maximizes the volume, insert it back into the volume equation to obtain the largest volume possible for the inscribed cone and round to two decimal places.