Optimization Problem:

Find the dimensions of the right circular cylinder of greatest volume inscribed in a right circular cone of radius 10" and height 24"

Draw a side view of the situation

(The cone will look like an isosceles triangle with a rectangle (the cylinder) sitting on its base and touching the sides)

let the radius of the cylinder be r and let the height of the cylinder be h
Look at the small right-angled triangle at the right
we can set up a ratio because of similar triangles
h/(10-r) = 24/10
10h = 240-24r
h = 24 - 12r/5

V = πr^2 h
= πr^2 (24 - 12r/5)
=24πr^2 - (12π/5) r^3

dV/dr = 48πr - (36π/5)r2
= 0 for a max/min

48πr = (36π/5)r^2
÷ by 12π
4r = 3π/5 r^2
divide by r, since r ≠ 0, it would give a minimum volume
4 = 3r/5
20 = 3r
r = 20/3
then h = 24 - (12/5)(20/3) = 8

the cylinder has a radius of 20/3 and a height of 8

check:
with those dimensions, V = 1117.01
let r = 19/3 (a little less) , then h = 44/5 , V = 1108.9 , a bit less
let r = 21/3 ( a little more), then h = 36/5 , V = 1108.35 (again, a little less)

My answer looks reasonable

To find the dimensions of the right circular cylinder of greatest volume inscribed in a right circular cone, we can use an optimization approach. The volume of a cylinder is given by V_cylinder = π * r^2 * h, where r is the radius of the cylinder, and h is the height of the cylinder. Additionally, there are some geometric constraints given for this problem:

1. The cylinder is inscribed in a cone, which means that the base of the cylinder (a circle) lies completely on the base of the cone.
2. The radius of the cylinder is equal to the radius of the base of the cone.
3. The height of the cylinder is less than or equal to the height of the cone.

Let's consider a cross-section of the cone and the inscribed cylinder. In this cross-section, let's assume that the center of the base of the cone coincides with the center of the base of the cylinder.

The radius of the base of the cone is given as 10", so the radius of the cylinder (r) is also 10".

The height of the cone is given as 24" and the height of the cylinder (h) is unknown.

To proceed with the optimization, we need to express h in terms of r so that we can write the volume of the cylinder as a function of a single variable (r).

From similar triangles, we can see that the height of the cone (24") and the height of the cylinder (h) form a proportion:

h/24 = r/10

By cross-multiplying, we get:

h = 2.4r

Now, substituting the expression for h in terms of r into the formula for the volume of a cylinder, we have:

V_cylinder = π * r^2 * (2.4r) = 7.2π * r^3

To maximize the volume, we can take the derivative of V_cylinder with respect to r and set it equal to zero, to find the critical points. The critical points will correspond to either a maximum or minimum volume.

dV_cylinder/dr = 7.2π * 3r^2

Setting this derivative equal to zero:

7.2π * 3r^2 = 0

Simplifying:

21.6π * r^2 = 0

r^2 = 0

Since r^2 cannot be zero, there are no critical points in the interior of the feasible region.

Next, we need to consider the endpoints of the feasible region. The radius, r, cannot be greater than the radius of the cone, which is 10". Therefore, the feasible domain for r is [0, 10].

Testing the endpoints:

When r = 0, V_cylinder = 0.
When r = 10, V_cylinder = 7.2π * 10^3 = 720π.

Comparing the volumes at the endpoints, we can see that the volume is greatest when r = 10". Therefore, the dimensions of the right circular cylinder of greatest volume inscribed in the right circular cone are:

Radius: 10"
Height: 2.4 * 10 = 24" (this comes from the proportion we derived earlier)

Hence, the cylinder has a radius of 10" and a height of 24".

To solve this optimization problem, we need to determine the dimensions of the right circular cylinder that will give us the maximum volume when inscribed in the given right circular cone.

Let's start by defining some variables:
- Let r be the radius of the circular base of the cylinder.
- Let h be the height of the cylinder.

Now, let's express the volume of the cylinder in terms of r and h. The volume of a right circular cylinder is given by V = πr²h.

To proceed with the optimization, we need to find a relationship between the radius r and height h of the cylinder.

Consider a cross-section of the cone and the inscribed cylinder. The radius of the circular base of the cone is 10 inches, and the height of the cone is 24 inches. We can use similar triangles to find the relationship between r and h.

The large triangle formed by the cone has side lengths 10, 24, and √(10² + 24²). The small triangle formed by the cylinder has side lengths r, h, and √(r² + h²).

By setting up the ratios of corresponding sides, we have:

r / h = 10 / 24 = 5 / 12

Now we can express the volume of the cylinder solely in terms of r:

V = πr²(5h/12)

Next, we can substitute the relationship between r and h into the volume equation to obtain an equation with a single variable. We'll substitute h with (12r/5):

V = πr²(5*(12r/5)/12)
= πr²(12r/12)
= πr³

Now we have the volume equation V = πr³.

To find the maximum volume, we can take the derivative of the volume equation with respect to r and set it equal to zero:

dV/dr = 3πr² = 0

Solving for r:

3πr² = 0
r² = 0
r = 0

However, r=0 is not a valid solution since it would result in a cylinder with no volume. Therefore, we need to consider the endpoints of the feasible interval.

The radius r cannot exceed the radius of the circular base of the cone, which is 10". Thus, we restrict our attention to the interval 0 ≤ r ≤ 10.

Evaluating the endpoints of the interval, we find that V = π(0)³ = 0 and V = π(10)³ = 1000π.

Therefore, the maximum volume of the right circular cylinder inscribed in the given right circular cone is 1000π cubic inches when the radius of the cylinder is 10 inches.