A right circular cone is inscribed in a sphere of radius r. Find the dimensions of the cone that maximize the volume of the cone.

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For a cone, the formula describes the relationship between the radius r of the base, the volume V, and the height h. Find the volume if the radius is 7 inches and the cone is 11 inches high. (Use 3.14 as an approximation for , and round to the nearest tenth.)

To find the dimensions of the cone that maximize the volume, we need to maximize the volume function in terms of the cone dimensions.

Let's start by drawing a diagram of the situation.

Since the cone is inscribed in a sphere, the height (h) of the cone will be equal to the diameter (2r) of the sphere.

Let r be the radius of the base and let h be the height of the cone.

From the diagram, we can see that the radius of the base of the cone (r), height of the cone (h), and the radius of the sphere (r) form a right triangle.

Using the Pythagorean theorem, we can write:

r^2 + h^2 = (2r)^2

Simplifying this equation, we get:

r^2 + h^2 = 4r^2

Subtracting r^2 from both sides, we have:

h^2 = 3r^2

Solving for h, we get:

h = sqrt(3r^2)

The volume of a cone is given by the formula:

V = (1/3) * π * r^2 * h

Substituting the expression for h in terms of r, we have:

V = (1/3) * π * r^2 * sqrt(3r^2)

To maximize the volume, we will differentiate V with respect to r and set it equal to zero to find the critical points.

dV/dr = (1/3) * π * (2r * sqrt(3r^2) + r^2 * (1/2) * (1/sqrt(3r^2))) = 0

Simplifying this equation, we have:

(2r * sqrt(3r^2)) + (r^2 / (2 * sqrt(3r^2))) = 0

Multiplying through by 2sqrt(3r^2), we get:

2r^2 + r^2 / (sqrt(3r^2)) = 0

Multiplying through by sqrt(3r^2), we have:

2r^2 * sqrt(3r^2) + r^2 = 0

2r^2 * sqrt(3r^2) = -r^2

Dividing through by -r^2, we get:

2sqrt(3r^2) = -1

Squaring both sides, we have:

4 * 3r^2 = 1

12r^2 = 1

Dividing through by 12, we get:

r^2 = 1/12

Taking the square root of both sides, we have:

r = 1 / sqrt(12) = sqrt(12) / 12

Thus, the radius of the base of the cone that maximizes the volume is sqrt(12) / 12.

To find the corresponding height, we can substitute this value of r back into the equation for h:

h = sqrt(3r^2) = sqrt(3*(sqrt(12)/12)^2)

Simplifying this expression, we get:

h = sqrt(3*(12/144))
h = sqrt(36/144)
h = sqrt(1/4)
h = 1/2

Therefore, the height of the cone that maximizes the volume is 1/2.

In summary, the dimensions of the cone that maximize the volume are:
- The radius of the base: sqrt(12) / 12
- The height of the cone: 1/2

To find the dimensions of the cone that maximize its volume, we need to set up an equation for the volume of the cone and then use calculus to find the optimal solution.

Let's start by considering the cone itself. We can define its height as "h" and its radius as "r_cone". We want to maximize the volume of the cone, which is given by the formula:

V_cone = (1/3) * π * r_cone^2 * h

Now, let's think about the relationship between the cone and the sphere it is inscribed in. Since the cone is inscribed in the sphere, the diameter of the sphere is equal to the height of the cone plus twice the radius of the cone. This can be expressed as:

2 * r_cone + h = 2r

Simplifying this equation, we get:

h = 2r - 2 * r_cone

Now, we can substitute this value of h into the formula for the volume of the cone:

V_cone = (1/3) * π * r_cone^2 * (2r - 2 * r_cone)

Next, let's express the volume of the cone as a function of a single variable. Since our goal is to find the dimensions of the cone that maximize its volume, we can express the volume as a function of the radius of the cone, r_cone:

V_cone(r_cone) = (1/3) * π * r_cone^2 * (2r - 2 * r_cone)

Now, we can take the derivative of V_cone with respect to r_cone and set it equal to zero to find the critical points:

dV_cone/dr_cone = 0

Differentiating the function V_cone(r_cone) with respect to r_cone, we get:

dV_cone/dr_cone = (1/3) * π * 2 * r_cone * (2r - 2 * r_cone) + (1/3) * π * r_cone^2 * (-2) = 0

Simplifying this equation, we have:

2 * r_cone * (2r - 2 * r_cone) - r_cone^2 = 0

Expanding, we get:

4r_cone - 4 * r_cone^2 - r_cone^2 = 0

Simplifying further, we have:

5 * r_cone^2 - 4r_cone = 0

Factoring out r_cone, we get:

r_cone(5 * r_cone - 4) = 0

Setting each factor equal to zero, we find there are two possible solutions:

1. r_cone = 0
2. 5 * r_cone - 4 = 0

The first solution, r_cone = 0, corresponds to a degenerate cone with zero volume and is not physically meaningful.

Solving the second equation for r_cone, we find:

5 * r_cone - 4 = 0
5 * r_cone = 4
r_cone = 4/5

Therefore, the dimensions of the cone that maximize its volume are a height of h = 2r - 2 * r_cone = 2r - 2 * (4/5) = 2r - 8/5, and a radius of r_cone = 4/5.