Study on the Maximum Volume of Inscribed Elliptical Cone inside a Cylinder

The volume of a cone is one third of the volume of a cylinder if they share same base and equal height. Let’s propose a hypothesis, if we expand the research area from cones to elliptical cones, could the maximum ratio of volume beyond one third? This paper tries to pull away the veils of it. Firstly, we present four types of inscribed elliptical cones inside cylinders, and consequently all the other inscribed elliptical cones can be classified to these four types. Secondly, for each type of them, this paper discusses the corresponding volume ratio of the inscribed elliptical cone to the cylinder. It is concluded that the largest ratio of the volume of an inscribed cone to that of the cylinder is 1/3. Finally, two types whose ratio could reach 1/3 are given as examples.


Introduction
Past learning tells us that the volume of a cylinder is three times that of the cone which shares same base and equal height with the cylinder. In turn, any cone is a third part of the cylinder which has the same base with it and equal height.
However, this is the simplest case. This paper extends inscribed cones to inscribed elliptical cones, under the condition of a fixed cylinder, to study and solve the volume of various types of inscribed elliptical cones, and determine which type of inscribed elliptical cones has the largest volume and if the maximum value is more than one third or not. In order to define the scope of the elliptic cones in this paper, we firstly provides the definition of inscribed elliptical cones: Definition of inscribed elliptical cones: If the bottom base of a cone is in-Type I Beginning with the simplest case, Figure 1 shows a cone and a cylinder in same base and equal height.

Type II
The inscribed elliptical cone is placed horizontally in the cylinder, and the in-

Lemma 1: Ellipse Area Theorem
The area of an ellipse is given by: e S ab π = where a and b are the semi-major and semi-minor axes of the ellipse. Proof: The standard equation for an ellipse centered at the origin is given by:

Solution of the Volume of the Inscribed Elliptical Cone
In this section, we solve the four types of inscribed elliptical volumes. For convenience, we convert the exact volume of the inscribed elliptical cone into the ratio to the cylinder.

Type I
Beginning with the simplest case, Figure 5 shows a cone and a cylinder in same base and equal height. S is the area of base, H is the height, R is the radius, moreover, a and b are the semi-major and semi-minor axes of the elliptical base.
By Lemma 2, it's easy to know that V c = 1/3πR²H, because a = b = R. Thus, Hence the ratio of volume is 1/3.

Type II
The cone is placed horizontally inside the cylinder, as shown in Figure 6, two intersection points to the upper and lower base are C, D, respectively, and the contact points to the side of cylinder are E, F, respectively. The figure of circle O is the aerial view of Figure 6.
For easier calculation, we set the directions of line CD and EF, respectively, as the directions of semi-major axis a and semi-minor axis b (The same as in the following cases). Because of the variety of the relationship between the height of the cylinder and the radius of the bottom base, it's possible that a < b, but this will not affect the volume of the elliptic cone. Now assuming that OC = x, H is the height, R is the radius, then we give the expression: According to the Lemma 2, Therefore,    Figure 8.
In addition, By the Lemma 2: Mark the volume ratio of the inscribed elliptical cone to the cylinder as T: Since the difficulty of directly solving the maximum value of this equation.
Hence, the variable substitution and the scaling method are chosen to solve the maximum value.

of Applied Mathematics and Physics
Let function Therefore, the function P(t) is monotonically decreasing on the interval (0,1/3), monotonic increasing on the interval [1/3, 1], In comparison of the value of function P(t) at 0 and 1, the maximum value can be obtained that is the P(1) = 1/3, so The criteria of this equality is: , the volume ratio of the inscribed elliptical cone to the cylinder is less than or equal to 1 3 .

Case 2: H < 2R
In this case, the cylinder is considered as a short and thick type. The two stages need to be considered at this moment:     In the first stage，the right triangle CMN in the front view of Figure 10, according to the Pythagorean Theorem: The front view of Figure 9, assume AC = x, By the Lemma 2: Assume the volume ratio of the ellipsoid to cylinder is T, then 1 4 Mark the right side of the inequality as function ( ) P t , 2R H as λ , obviously 1 λ > . In order to solve the maximum value of the function ( ) P t , we take the derivative of the function ( ) P t , so: '( ) 0 P t ≥ . Therefore, the function ( ) P t is initially increasing, but turns to decrease after. Since , thus the maximum volume ratio of the cone is The equality doesn't hold actually, which means the sign of the less than is strictly true. When H < 2R, at the first stage, the maximum volume ratio of the inner elliptical cone is less than 1/3.
Let research the maximum volume ratio of the inscribed elliptical cone in the second stage, at present, , as shown in Figure 11.
Rearrange to get: By the Lemma 2: 1 / 3* * * * / 1 / 3* * 1 * * 3 4 4 * * * 4 Mark the right side as function ( ) P t , Again in order to solve the maximum value of the function ( ) P t , we take the derivative of the function ( ) P t , so: Figure 3 (3-1)), thus, the function ( ) P x is initially increasing, but turns to decrease after, the When the cone is in the stage as shown in Figure 12, the maximum volume ratio is achieved, but still less than 1/3, that means the maximum volume ratio of the inscribed elliptical cone to the cylinder in the second stage is less than 1/3.
Summarizing, when H < 2R, the maximum volume ratio of the inner elliptical cone to the cylinder is strictly less than1/3.

Type IV
The x y a + = , and the semi-minor axis    x y ≥ , which can also be proved by the geometric theory based on Figure 13.
Thus, the volume ratio of the inscribed elliptical cone to the cylinder is: (2 / 2) * 2 1 / 3* * * 1 / 3* * * * ( , ) * * *  Rx , if he vertex of the inscribed elliptical cone is exact on the point N, then the volume of the cone is less than or equal to one third the volume of the cylinder, and the solely criteria holds the equal sign is that 2) The vertex of the inner elliptical cone falls on the side of the cylinder Since ∆ADC ∽ ∆PMN, thus In addition, Then, calculate the volume ratio of the inscribed elliptical cone to the culinder by the Lemma 2: In order to ensure that the vertex of the inscribed elliptical cone falls on the side of the cylinder, letting To check the variation of 1 ( , ) T x y with y, we calculate the partial derivative of 1 ( , ) T x y by y: . Therefore, 1 ( , ) T x y will decrease first and then increase with the increase of y. Thus we get:  H) is at its maximum, which can be classified as the first situation of type III that is when H ≥ 2. Now the volume ratio of the inscribed cone to the cylinder is less than or equal to one third, and the equal sign holds when H = 2R, x = 2R.
In summary, in type IV, when H ≥ 2R, if the vertex of the inscribed elliptical cone falls on the side of the cylinder, the ratio of the volume of the cylinder is less than or equal to 1/3, and the equal sign holds when: Furthermore, as According to the Lemma 2, the ratio of the volume of the elliptical cone to the volume of the cylinder is:   x M x y ≥ , i.e. the variable x is monotone increasing, therefore: In order to study the variation of (2 , ) M R y with the change of y, calculate the partial derivative of (2 , ) M R y with y: Since the range of y is  x y a + = , and the semi-minor axis In order to study the variation of ( , ) T x y with the change of y, we calculate its partial derivative of y： Therefore, when H < 2R and the vertex of the inscribed elliptical cone falls just on the bottom of the cylinder, the ratio of the volume is less than 1/3.
1) The vertex falls on the side of the cylinder Figure 17 shows the situation when the vertex falls on the side of the cylinder.
In the front view of Figure    x y According to the Lemma 2, the volume ration of the elliptical cone to the cy- Hence, when H < 2, y = H, the volume ratio reaches its maximum, i.e. And this situation can be classified in the discussion when H<2R, thus:  The variation of ( , ) T x y is also the same as the first situation when 4 3 2 H R ≤ , that is the ratio of the inscribed elliptical cone volume to the volume of the cylinder is less than or equal to one third , and the equal sign holds when 2 , 0 x R y = = .
Note: we need to point out that in case 3 and case 4, this paper gives different examples of the inscribed elliptical cone, whose bottom oval area is a possible maximum value. This means that, in some cases, the maximum area of the ellipse cannot reach the value that is given in this paper, but this won't influences the conclusion .In addition, the maximum area of the oval can be reached when 2 , 0 x R y = = .

Summary
In this paper, we studied the maximum volume ratio of the inscribed elliptical cone in four different situations, 1) the most common case when the volume ration is 1/3; 2) the volume ration is strictly less than 1/3; 3) the volume ratio is less than or equal to 1/3, and the equal sign holds when 2 , 2 H R x R = = , (This is a rather special case); 4) the volume ratio is less than or equal to 1/3, and the equal sigh holds when

Future Study
In this paper, we studied the maximum volume ratio of the inscribed elliptical cone to the cylinder. In the future study, we can expand the research area from cylinder to elliptical cylinders, cuboids and prisms. With the similar method, we can study the maximum volume ratio of the inscribed elliptical cone and the inscribed pyramid.