How to find the volume of a parallelepiped?

Find the angle of a wedge of a circle that produces the cone with the greatest volume?

• calculus activity. PLEASE HELP part 1: cut a wedge from a circle and remove it. form the remaining piece of the circle into the cone. find the angle of wedge that produces the cone with the greatest volume. part 2: make a second cone from the removed wedge. find a formula for the volume of this second cone in terms of theta, the angle of the wedge. part 3: find the wedge angle that produces the maximum total volume of the two cones

• Answer:

  I'll get you started. Since the radius of the circle is really immaterial, we can assume without loss of generality that it is 1. The circumference of the circle is of course C_1 = 2π. Now, consider removing a sector with central angle Θ. The arclength of this sector is Θ times the radius of the circle. With radius 1, the sector arclength is S = Θ. Fold the edges together to get a cone. The circumference of the cone is 2π - S = 2π - Θ. If the base radius of the cone is r, then 2πr = 2π - Θ ==> r = 1 - Θ/2π. The radius of the circle, 1, is the slant height of the cone. The height and radius of the cone form a right triangle with the slant height. Hence the height of the cone h is h = √(1² - r²) = √(1 - 1 + Θ/π - Θ²/4π²) = 1/(2π) √(4πΘ - Θ²). The volume of the cone can be written as a function of Θ. V = (π/3) r² h = (π/3) (1 - Θ/2π)² 1/(2π) √(4πΘ - Θ²) ==> V(Θ) = 1/(24π²) (2π - Θ)²√(4πΘ - Θ²), for 0 < Θ < 2π. Now take the derivative, find the critical value(s) and verify that you get a maximum. You'll find two critical values, but only one makes sense because Θ < 2π. The critical value is Θ = 2π(1 - √(2/3)). The maximum volume for this cone is V_max = 16√(3)π^3/9. The circumference of the smaller cone is just Θ, the arclength of the removed wedge. The radius R of the smaller cone satisfies 2πR = Θ ==> R = Θ/2π. The height H and radius of the smaller cone still make a right triangle with the radius 1 of the circle. So the smaller cone has height H = √(1 - R²) = √(1 - Θ²/4π²) = 1/(2π) √(4π² - Θ²). The smaller cone has volume v = (π/3)R²H = (π/3) (Θ/2π)² 1/(2π) √(4π² - Θ²) = 1/(24π²) Θ² √(4π² - Θ²). For part 3, find the maximum value of the combined volumes V_T(Θ) = V + v = 1/(24π²)[(2π - Θ)²√(4πΘ - Θ²) + Θ² √(4π² - Θ²)].