Constructing a Cone with an Optimized Volume By Mr. E Calculus (year 2003) b

The Materials Needed Poster Board Compass Protractor Scizzors Ruler Tape Staples Colored Pencils Calculator Computer

Procedure 1. Using a Compass, and ruler, trace a Circle with a radius of 4 inches. 2. Cut the Circle out of the Poster Board 3. Repeat steps 1 and 2 two more times 4. Use the ruler and protractor to cut two of the circles with 45º and 60º slices, respectively. Label them Cone 45º and Cone 60º, respectively. 5. Form the cones from the slices in the circle by joining the ends and either stapling or taping them together.

Inquiry/Question Which of the two Cones contains largest volume? Take a quess. To find out, measure the radius of the base circle and the height of the cones using the ruler. Use the formula Volume=1/3*  r²h and calculate the Volume of Cone 45º and Cone 60º, and compare the answers.

Hypothesis Is it possible to construct a cone which will contain the maximum Volume?

Data The Circumference of a circle is determined by the formulu C=  d The height of the cone lives in the relationship: h² + r²= R² where h is the height of the cone, r is the radius of the cone’s base circle, and R is the radius of the circle which was originally used to construct the cone( see the diagram) The Volume of a Cone is calculated by using the formula: Volume=1/3*  r²h

Diagrams R X R h r The Original Circle to be cut with arc length X and radius R The completed cone will have a height of h, a base radius of r, and a face diagonal of R.

Diagrams R X The Original Circle to be cut with arc length X and radius R

Diagrams R The Original Circle to be cut with arc length X and radius R Close the ends together

Diagrams R The Original Circle to be cut with arc length X and radius R Close the ends together

Diagrams R. Pull top up to form cone

Diagrams R h r The completed cone will have a height of h, a base radius of r, and a face diagonal of R. Pull top up to form cone

Diagrams R h r The completed cone will have a height of h, a base radius of r, and a face diagonal of R.

Analysis and Exploration Note that the cone Volume will be dependent on the size X arc length which will be cut away Note that the arc length X determines the angle that is cut. Hence the angle cut from the original circle determines the Volume of the cone The Volume of the cone can be optimized by determining the X that can be cut yielding the greatest Volume by using Calculus or Trial and Error.

Trial and Error Cone Volume Optimization Keep constructing different cones by cutting different angles away from the original circle of posterboard. This process could take several hours,…or one could…

Use the Calculus Approach to maximize the Volume of a Cone Step 1: Determine the first derivative by differentiating the Volume formula with respect to X. Volume= V(x) = 1/3*  r²h dV/dx = 1/3*  * [r²*dh/dx + h* 2*r*dr/dx]

Use the Calculus Approach to maximize the Volume of a Cone Step 2: Alter the formula for the Circumference C=  d =  (2R)=  (2*4)=8  The circumference will be reduced by X in order to construct the cone. The new cone will have a base circle circumference of: 2  r = (8  - x). Therefore the base radius r formula will be r = (8  - x)/ (2  ) = 4 – (x/(2  ))

Use the Calculus Approach to maximize the Volume of a Cone Step 3: Determine the first derivative by differentiating the radius r with respect to X. r = (8  - x)/ (2  ) = 4 – (x/(2  )) dr/dx = 0 – 1/2  dr/dx = -1/2 

Use the Calculus Approach to maximize the Volume of a Cone Step 4: Recall that the height of the cone lives in the relationship: h² + r²= R² where h is the height of the cone, r is the radius of the cone’s base circle, and R is the radius of the circle which was originally used to construct the cone( see the diagram) Performing implicit differentiation on h² + r²= R² yields 2h dh/dt + 2r dr/dt = 0 since R is a constant Solving for dh/dt = -2r/2h* dr/dt = -r/h*(-1/2  )= = r/(2  h)

Use the Calculus Approach to maximize the Volume of a Cone Step 5: Substitute dh/dx into the dV/dx derivative. Volume= V(x) = 1/3*  r²h dV/dx = 1/3*  * [r²*r/(2  h) + h* 2*r*dr/dx] dV/dx = 1/3*  *[r 3 /(2  h) + h*2*r*(-1/ 2  )] dV/dx = 1/3*  * [r 3 /(2  h) - h*r*/(  )] Set dV/dx = 0 to maximize the Volume 0 =1/3*  * [r 3 /(2  h) - h*r*/(  )] Mulitplying both sides by 3/  0 = [r 3 /(2  h) - h*r*/(  )] Multiplying everything by 2  h 0 = r 3 – 2h²r Adding 2h²r to both sides, 2h²r = r 3 Dividing both sides by r 2h²= r 2

Substituting the value of h and the value of r: 2(16 – (8  - x) 2 / (2  ) 2 ) = ((8  - x)/ (2  )) 2 multiplying both sides by 4   (8  - x) 2 = (8  - x) 2 Adding 2 (8  - x) 2 to both sides 128  2 = 3(8  - x) 2 Dividing both sides by 3 And getting the square root of both sides (8  - x) =  √128/3 Solving for x we get x = 8  -  √ 42.67=  *(8-6.53) =  *1.47= 4.62

Conclusion X= 4.62 will yield the Cone with the largest Volume. Inguiry: What angle is cut when the arc length x=4.62? (use the Protractor to find out)

Max Volume of Cone

Volume of Cone (Excel Chart) The Max Volume = at x=4.62

Now you may color your Cones!!! Use the colored pencils to color the cones with your favorite designs