Infinitely many different sectors can be cut from a circular piece of paper with a 12-cm radius, and any such sector can be fashioned into a paper cone with a 12-cm slant height.
(a) Show that the volume of the cone produced by the 180-degree sector is larger than the volume of the cone produced by the 120-degree sector.
(b) Find a sector of the same circle that will produce a cone whose volume is even larger.
(c) Express the volume of a cone formed from this circle as a function of the central angle of the sector used to form it, then find the sector that produces the cone of greatest volume.
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(a) To show that the volume of the cone produced by the 180-degree sector is larger than the volume of the cone produced by the 120-degree sector, we can compare their volumes directly.
The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius of the base and h is the height or slant height of the cone.
The 180-degree sector will have a central angle of 180°, which means the cone produced will have a height equal to the radius of the circle (12 cm) and a base radius equal to the slant height of the cone.
Using the Pythagorean theorem, the slant height s of the cone can be found as:
s² = r² + h²
s² = (12 cm)² + (12 cm)²
s² = 144 cm² + 144 cm²
s² = 288 cm²
s = √(288 cm²)
s ≈ 16.97 cm
The volume V₁ of the cone produced by the 180-degree sector is given by:
V₁ = (1/3)π(s₁)²h₁
V₁ = (1/3)π(16.97 cm)²(12 cm)
V₁ ≈ 305.91 cm³
The 120-degree sector will have a central angle of 120°, which means the cone produced will have a height equal to half the radius of the circle (6 cm) and a base radius equal to the slant height of the cone.
Using the Pythagorean theorem, the slant height s₂ of the cone can be found as:
s₂² = r² + h₂²
s₂² = (12 cm)² + (6 cm)²
s₂² = 144 cm² + 36 cm²
s₂² = 180 cm²
s₂ = √(180 cm²)
s₂ ≈ 13.42 cm
The volume V₂ of the cone produced by the 120-degree sector is given by:
V₂ = (1/3)π(s₂)²h₂
V₂ = (1/3)π(13.42 cm)²(6 cm)
V₂ ≈ 351.86 cm³
Comparing the volumes, V₁ ≈ 305.91 cm³ and V₂ ≈ 351.86 cm³, we can see that V₂ is larger than V₁. Therefore, the volume of the cone produced by the 180-degree sector is smaller than the volume of the cone produced by the 120-degree sector.
(b) To find a sector of the same circle that will produce a cone with an even larger volume, we need to increase the central angle of the sector.
Let's consider a sector with a central angle of 240°. The cone produced by this sector will have a height equal to two-thirds of the radius of the circle (8 cm) and a base radius equal to the slant height of the cone.
Using the Pythagorean theorem, the slant height s₃ of the cone can be found as:
s₃² = r² + h₃²
s₃² = (12 cm)² + (8 cm)²
s₃² = 144 cm² + 64 cm²
s₃² = 208 cm²
s₃ = √(208 cm²)
s₃ ≈ 14.42 cm
The volume V₃ of the cone produced by the 240-degree sector is given by:
V₃ = (1/3)π(s₃)²h₃
V₃ = (1/3)π(14.42 cm)²(8 cm)
V₃ ≈ 548.14 cm³
Comparing the volumes, V₂ ≈ 351.86 cm³ and V₃ ≈ 548.14 cm³, we can see that V₃ is larger than V₂. Therefore, the sector with a central angle of 240° will produce a cone with an even larger volume.
(c) To express the volume of a cone formed from this circle as a function of the central angle of the sector used to form it, let's use the symbol θ to represent the central angle.
The height h₄ of the cone can be determined as h₄ = r * sin(θ/2), where r is the radius of the circle.
The slant height s₄ can be determined using the Pythagorean theorem as s₄² = r² + h₄².
Substituting the expression for h₄ into the equation for s₄, we get s₄² = r² + (r * sin(θ/2))².
The volume V₄ of the cone formed from this sector is given by:
V₄ = (1/3)π(s₄)²h₄
V₄ = (1/3)π(r² + (r * sin(θ/2))²)(r * sin(θ/2))
To find the sector that produces the cone of greatest volume, we need to find the value of θ that maximizes the volume function V₄. The easiest way to do this is by taking the derivative of V₄ with respect to θ, setting it equal to zero, and solving for θ.
To answer these questions, we need to understand how to calculate the volume of a cone and how it is related to the central angle of the sector used to form it.
(a) To compare the volumes of the cones produced by the 180-degree and 120-degree sectors, we can use the formula for the volume of a cone: V = (1/3)πr²h, where r is the radius of the base and h is the height (or slant height) of the cone.
The 180-degree sector covers half of the circle, so the radius of the base of the cone is 12 cm, and the slant height is also 12 cm (given in the problem). Plugging these values into the volume formula: V₁ = (1/3)π(12)²(12) = 576π cm³.
The 120-degree sector covers one-third of the circle, so the radius of the base of the cone is still 12 cm, but the slant height is now unknown. We can use trigonometry to find the slant height: h = 2r*sin(θ/2), where θ is the central angle. For 120 degrees: h = 2(12)sin(120/2) ≈ 20.78 cm. Plugging these values into the volume formula: V₂ = (1/3)π(12)²(20.78) ≈ 994.76π cm³.
Comparing V₁ and V₂, we find that V₂ > V₁, so the volume of the cone produced by the 120-degree sector is larger than the volume of the cone produced by the 180-degree sector.
(b) To find a sector that produces an even larger cone, we need to increase the central angle. By taking a sector with a larger central angle, we cover more of the circle, resulting in a larger base radius and potentially a larger slant height.
If we take a sector that covers the entire circle (360 degrees), the radius of the base will be 12 cm, and the slant height will also be 12 cm (given in the problem). Plugging these values into the volume formula: V₃ = (1/3)π(12)²(12) = 576π cm³.
Comparing V₃ with V₂, we find that V₃ = V₁, indicating that the volume of the cone produced by the 360-degree sector is the same as the volume of the cone produced by the 180-degree sector. Thus, any sector that covers the entire circle will produce a cone with the same volume.
(c) To express the volume of a cone formed from this circle as a function of the central angle, we can use the radius (r) and the slant height (h) as variables. The volume formula becomes V = (1/3)πr²h.
Since the slant height (h) is related to the central angle (θ) by h = 2r*sin(θ/2), we can substitute this expression for h into the volume formula. We get: V = (1/3)πr²(2r*sin(θ/2)) = (2/3)πr³*sin(θ/2).
Now, to find the sector that produces the cone of the greatest volume, we need to find the maximum value of the function V = (2/3)πr³*sin(θ/2).
To maximize this function, we take the derivative of V with respect to θ, set it equal to zero, and solve for θ. This will give us the critical angle(s) where the volume is maximum.
dV/dθ = 0
(2/3)πr³*(1/2)*cos(θ/2) = 0
cos(θ/2) = 0
The critical angles occur where cos(θ/2) = 0, which is when θ/2 = π/2 or θ/2 = 3π/2. Solving for θ, we find θ = π or θ = 3π.
So, the sectors that produce the cone of the greatest volume have central angles of π or 3π.
Note: Please double-check the calculations and final answers as this is an explanation of the steps involved in solving the problem.