Maximum+Volume+of+a+Cone

A cone is constructed from a circle by removing a sector and then connecting two radii. The volume of this cone depends on the size of the sector removed. //What is the volume of the largest cone that can be constructed from a circular piece of paper with a given diameter? What is the arc length of the removed sector?What is the central angle of the removed sector? What patterns do you notice?//

__Mathematics Behind Maximum Volume Formulas__
The volume of a cone is represented by the following equation: math V=\frac{\pi}{3}r^{2}H math

We want to maximize this volume, and we can do so by using the "fmax" function on our calculators. In order to use the "fmax" function in our calculator program we first need to rewrite our volume equation in terms of a single variable. I will show the steps below for creating a volume equation in terms of the variable H and the constant, R.

Looking at the right triangle inside our cone, we have math H=\text{height}\\\ r=\text{radius of cone base}\\\ R=\text{hypotenuse of cone} math

R is determined by the diameter of our given circle. math R=\frac{D}{2} math

I want to solve for r in terms of variable,H, and constant,R. I will do this by using the Pythagorean Theorem. math r^{2}+H^{2}=R^{2}\\ r^{2} =R^{2}-H^{2}\\\ r=\sqrt{R^{2}-H^{2}} math

Now we can substitute r into our volume equation. math V=\frac{\pi}{3}(\sqrt{R^{2}-H^{2}})^{2}H\\ V=\frac{\pi}{3}(R^{2}-H^{2})H\\ V=\frac{\pi}{3}(R^{2}H-H^{3})\\ math

Our volume equation is now in terms of a single variable,H, and a single constant, R. We can now use the "fmax" function to find the maximum height of a cone given a specific diameter. The input for the "fmax" function is as follows: math fmax(\text{expression, variable, left bound, right, bound})\\ fmax(\frac{\pi}{3}(R^{2}H-H^{3})), H, 0, D) math This function provides us with the maximum height (stored as M) that will produce the maximum volume given a diameter, D.

Now we need to look at the cone with the maximum height, to find the radius of the arc sector.

This right triangle has a hypotenuse R and a height of M. Using the Pythagorean theorem, we can solve for the length of the arc sector, S. math M^{2}+S^{2}=R^{2}\\ S^{2} =R^{2}-M^{2}\\\ S=\sqrt{R^{2}-M^{2}} math

To find the maximum volume of a cone, given a certain diameter, D, we substitute M and S into our volume equation. math \text{Max Volume (V)}=\frac{\pi}{3}S^{2}M math

The table below provides further details of the code needed for a calculator program to can find the maximum volume of a cone using the 'fmax" function and a given diameter, D.

__Mathematics Behind Arc Length of Removed Sector Formulas__
Since we found the radius length of the sector removed above, we can find the arc length by substituting S into the arc length formula below. math A=D\pi-2{\pi}S\\ A=\text{arc length}\\ D=\text{diameter given}\\ S=\text{radius length of sector removed} math In the above formula, we find the arc length of the removed sector by subtracting the circumference of the circle with the arc sector from the circumference of the original circle.

__Mathematics Behind Central Angle of Removed Sector Formulas__
To find the central angle of the removed sector in degrees, we use the formula math A=D{\pi}(C/360)\\ \text{where A=arc length}\\ D=\text{diameter given}\\ C=\text{central angle} math Solving this formula for C gives the following equation for the central angle in degrees. math C=\frac{360A}{D\pi} math

Disp "DIAMETER OF" Disp "THE CIRCLE" PROMPT D || To begin the program, we display a question that prompts the user to input D, the diameter of the given circle. || Disp "IS" Disp V || This displays the maximum volume of the cone after the calculations have been made. || Disp "OF REMOVED" Disp "SECTOR IS" Disp A || This displays the arc length of the removed sector after the calculations have been made. || Disp "OF REMOVED" Disp "SECTOR IS" Disp C || This displays the central angle of the removed sector after the calculations have been made. ||
 * = === Code for Max Volume Calculator Program === ||
 * = **Code** ||= **Explanation** ||
 * Disp "WHAT IS THE"
 * D/2 -> R || The diameter is then divided by 2. Stored as "R". ||
 * fmax((pi/3)*(R^2*H-H^3), H,0,D)->M || When we enter an expression, the variable, left bound, and right bound into the fmax function, it will calculate the maximum of the expression. By entering the volume equation in terms of height, the fmax function found the height that would produce the maximum volume of the cone. The height is stored as "M". ||
 * sqrt(R^2-M^2) ->S || With the max height, M, and the large radius, R, we can use the Pythagorean theorem to solve for S, the length of the arc sector, which is the new radius once the arc sector has been removed. This cone radius is stored as "S". ||
 * D*(pi)-2*(pi)*S ->A || To find the arc length of the removed sector, we subtract the circumference of the cone base from the circumference of the original circle. The arc length of the removed sector is stored as "A". ||
 * (360*A)/(D*(pi)) ->C || To find the central angle in degrees, we multiply the arc length of the removed sector, A, by 360 degrees and then divide the value by the circumference of the original circle, D*pi. ||
 * (pi/3)*S^2*M ->V || Since we stored our maximizing values as M and S, we can input this values into the volume equation to produce the max volume. ||
 * Disp "MAX VOLUME" Disp "OF CONE"
 * Disp "ARC LENGTH"
 * Disp "CENTRAL ANGLE"

of Cone ||= Length of Arc Sector ||= Central Angle of Sector Removed ||
 * = Diameter of Circle ||= Max Volume
 * = 6 cm ||= 10.8828 cm^3 ||= 3.4590 cm ||= 66.0612 degrees ||
 * = 7cm ||= 17.2815 cm^3 ||= 4.0354 cm ||= 66.0612 degrees ||
 * = 9 cm ||= 36.7294 cm^3 ||= 5.1884 cm ||= 66.0612 degrees ||
 * = 20 cm ||= 403.0665 cm^3 ||= 11.5299 cm ||= 66.0612 degrees ||

__Patterns__
A noticeable pattern in above table is the equality of the central angles. Regardless of the diameter of the circle, every cone at its max volume with the given diameter will have the same central angle of 66.0612 degrees.

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