The slant height of a cone is square root 12cm. Find the maximum volume of the cone.
To find the maximum volume of a cone, we need to maximize the dimensions of the cone. Since the slant height is given as √12 cm, we can use this information to determine the height and radius of the cone.
Let's assume the height of the cone as h and the radius of the cone as r.
Using the Pythagorean theorem, we know that the slant height (l) is related to the height (h) and the radius (r) by the equation:
l^2 = h^2 + r^2
Given that the slant height (l) is √12 cm, we can substitute it into the equation:
(√12)^2 = h^2 + r^2
12 = h^2 + r^2 ...(1)
Now, the volume of a cone (V) is given by the formula:
V = (1/3) * π * r^2 * h
To maximize the volume, we need to express the volume in terms of a single variable. Solving equation (1) for r^2, we get:
r^2 = 12 - h^2
Substituting this into the formula for volume, we have:
V = (1/3) * π * (12 - h^2) * h
Now, let's find the value of h that maximizes the volume by finding the critical points.
To maximize or minimize a function, we need to find its derivative and set it equal to zero.
Let's differentiate V with respect to h:
dV/dh = (1/3) * π * (12 - h^2) - (1/3) * π * h * 2h
= (1/3) * π * (12 - 3h^2 - 2h^2)
= (1/3) * π * (12 - 5h^2)
Now, set dV/dh equal to zero and solve for h:
(1/3) * π * (12 - 5h^2) = 0
12 - 5h^2 = 0
5h^2 = 12
h^2 = 12/5
h = √(12/5) or h = -√(12/5)
Since the height of a cone cannot be negative, we can discard the negative value.
Therefore, the height of the cone that maximizes the volume is h = √(12/5) cm.
Substituting this value of h back into equation (1), we can find the value of r:
r^2 = 12 - h^2
r^2 = 12 - (12/5)
r^2 = 60/5 - 12/5
r^2 = (60 - 12)/5
r^2 = 48/5
r = √(48/5) cm
Now, we have the height (h) and radius (r) of the cone. We can substitute these values into the formula for the volume of the cone to find the maximum volume:
V = (1/3) * π * r^2 * h
V = (1/3) * π * (√(48/5))^2 * √(12/5)
V = (1/3) * π * (48/5) * √(12/5)
V = (1/3) * π * 48/5 * √(12/5)
V = (16/5) * π * √(12/5)
Therefore, the maximum volume of the cone is (16/5) * π * √(12/5) cubic cm.
To find the maximum volume of a cone given the slant height, we need to determine the radius and height that will maximize the volume. Let's break down the steps to solve the problem.
1. We are given the slant height of the cone, which is √12 cm. The slant height (l) relates to the radius (r) and height (h) of the cone using the Pythagorean theorem: l^2 = r^2 + h^2. In this case, we have (√12)^2 = r^2 + h^2.
2. Simplifying (√12)^2, we get 12 = r^2 + h^2.
3. To find the maximum volume, we need to express the volume of the cone (V) in terms of a single variable. The formula for the volume of a cone is V = (1/3) x π x r^2 x h.
4. We can express the volume (V) solely in terms of h by using the equation from step 2. Rearrange equation 2 to get h^2 = 12 - r^2. Substitute this into the formula for V to get V = (1/3) x π x r^2 x (12 - r^2).
5. Now, we have the volume of the cone as a function of r: V(r) = (1/3) x π x r^2 x (12 - r^2).
6. To find the maximum volume, we need to find the critical points of the function V(r). We can do this by taking the derivative of V with respect to r, setting it equal to zero, and solving for r.
7. Differentiate V(r) with respect to r, which involves using rules of differentiation and the product rule: dV/dr = (1/3) x π x [2r(12 - r^2) + r^2(-2r)].
8. Simplify the derivative: dV/dr = (1/3) x π x [24r - 4r^3 - 2r^3].
9. Set the derivative equal to zero and solve for r: 24r - 4r^3 - 2r^3 = 0.
10. Simplify the equation: -6r^3 + 24r = 0.
11. Factor out r: r(-6r^2 + 24) = 0.
12. Set each factor equal to zero and solve for r: r = 0, -6r^2 + 24 = 0.
13. From the equation -6r^2 + 24 = 0, solve for r: -6r^2 = -24, r^2 = 4, r = ±2. Since r cannot be negative in this context, we take r = 2.
14. Now that we have the value of r, we can substitute it back into the equation from step 2 to find h: 12 = 2^2 + h^2, 12 = 4 + h^2, h^2 = 8, h = √8 = 2√2.
15. We have obtained the values for r and h that maximize the volume. Plug these values into the formula for the volume of the cone to find the maximum volume.
V(maximum) = (1/3) x π x (2^2) x (2√2) = (4/3)π√2 cubic cm.
Therefore, the maximum volume of the cone is (4/3)π√2 cubic cm.