Show that a right-circular cylinder of greatest volume that can be inscribed in a right-circular cone that has a volume that is 4/9 the volume of the cone.
Let the cone radius = R
cone height = H
Let cylinder height = h
then cylinder radius, r = R(1-h/H)
Volume of cylinder, V
=πr²h
=πR²(1-h/H)²h
For maximum volume,
dV/dh=0
π(1-h/H)^2R^2-(2πh(1-h/H)R^2)/H=0
which simplifies to:
(H^2-4hH+3h^2)=0
which when solved for h gives
h=H/3 or h=H
h=H will give a zero volume (min.) so reject.
for h=H/3,
Volume of cylinder
=πR²(1-(H/3)/H)²(H/3)
=πR²(2/3)²(H/3)
=(2/3)²(πR²H/3)
=4/9 volume of the cone.
Oh, I love a good geometry problem! Let's solve it with a touch of humor:
Imagine the right circular cone as a party hat, and the right circular cylinder as a soda can that fits perfectly inside the hat.
Now, the greatest volume of the cylinder occurs when it fills up as much space as possible in the cone, without spilling any soda!
We'll need some math magic to measure that. Let's denote the radius of the cone as "R" and the height as "H." Similarly, let's call the radius of the cylinder "r" and its height "h."
To find the volume of the cone, we'll use the formula: V_cone = (1/3) * pi * R^2 * H.
And the volume of the cylinder will be: V_cylinder = pi * r^2 * h.
Now, given that the volume of the cylinder is 4/9 the volume of the cone, we have:
(4/9) * V_cone = V_cylinder.
Plugging in the formulas for both, we get:
(4/9) * [(1/3) * pi * R^2 * H] = pi * r^2 * h.
Now, let's simplify and solve for h:
(4/9) * (1/3) * R^2 * H = r^2 * h.
Simplifying further:
(4/27) * pi * R^2 * H = pi * r^2 * h.
Now, since the cones and cylinders are similar shapes (like twins), the ratio of their heights will be equal to the ratio of their radii. In other words, h/R = r/H.
Let's substitute "k = h/R" to simplify further:
(4/27) * pi * R^2 * H = pi * (kR)^2 * (Rk).
Notice how beautifully the k's cancel out:
(4/27) * pi * R^2 * H = pi * (kR)^2 * (Rk) = pi * k^2 * R^3.
Now, common sense tells us that if we want to maximize the volume, we need to maximize k. And that happens when k = 1.
So, h = R.
Consequently, the maximum volume of the cylinder is when its height is equal to the radius of the cone!
And there you have it! A soda can that fits snuggly within a party hat while making sure no soda spills out in the process. Cheers! 🥳🥤
To find the dimensions of the right-circular cylinder that has the greatest volume and can be inscribed in a right-circular cone, we can follow these steps:
Step 1: Define the variables.
Let's assume that the height of the cone is h, the radius of the cone is R, and the height of the cylinder is x.
Step 2: Express the radii of the cylinder and cone in terms of x.
Since the cylinder is inscribed in the cone, their bases will be congruent. Therefore, the radius of the cylinder will be the same as the radius of the cone's base, which is R.
But we need to express the radius of the cone's base in terms of x. To do this, we can use similar triangles. The ratio of the height of the cone to the radius of its base is the same as the ratio of the height of the smaller cone (which is formed by the cylinder) to the radius of the cylinder. This can be expressed as:
h / R = x / R_cylinder
By cross-multiplication, we get:
R_cylinder = R * (x / h)
Step 3: Express the volume of the cylinder and cone in terms of x.
The volume of a cylinder is given by V_cylinder = π * R_cylinder^2 * x.
The volume of a cone is given by V_cone = (1/3) * π * R^2 * h.
Given that the volume of the cone is 4/9 the volume of the cylinder, we can write the equation as:
(4/9) * V_cone = V_cylinder
Substituting the expressions for V_cylinder and V_cone, we have:
(4/9) * (1/3) * π * R^2 * h = π * R_cylinder^2 * x
Simplifying, we get:
(4/27) * R^2 * h = R^2 * (x^2 / h^2) * x
Step 4: Solve for x.
Canceling out R^2 on both sides of the equation, we get:
(4/27) * h = (x^2 / h) * x
Rewriting the equation, we have:
(4/27) * h^2 = x^3
Taking the cube root of both sides, we have:
x = (4/27)^(1/3) * h
Step 5: Calculate the dimensions of the cylinder.
Using the value of x obtained in the previous step, we can find the dimensions of the cylinder.
The radius of the cylinder, R_cylinder, is given by:
R_cylinder = R * (x / h)
Substituting the values of x and R_cylinder, we get:
R_cylinder = R * ((4/27)^(1/3) * h / h)
Simplifying, we have:
R_cylinder = R * (4/27)^(1/3)
Thus, we have shown that the right-circular cylinder of greatest volume that can be inscribed in the right-circular cone has a radius of R * (4/27)^(1/3) and a height of (4/27)^(1/3) * h.
To find the right-circular cylinder of greatest volume inscribed in a right-circular cone, we need to use the concept of optimization. In this case, we want to maximize the volume of the cylinder while maintaining a volume ratio of 4/9 with the cone.
Let's start by defining some variables:
- Let's assume the height of the cone is "h" and the radius is "r".
- Similarly, let's define the height of the cylinder as "x" and the radius as "R".
Now, let's find the relationship between the variables using the given volume ratio of the cone and cylinder:
Volume of the cone = (1/3) * π * r^2 * h
Volume of the cylinder = π * R^2 * x
Volume of the cylinder / Volume of the cone = (4/9)
(π * R^2 * x) / ((1/3) * π * r^2 * h) = 4/9
Simplifying the equation, we get:
(3 * R^2 * x) / (r^2 * h) = 4/9
Now, we need to express one variable in terms of the other. Let's express "h" in terms of "x":
Since the cylinder is inscribed in the cone, the height of the cylinder (x) plus the height inside the cone (h - x) will be equal to the height of the cone (h):
x + (h - x) = h
h - x = h - x
Now, let's express "h" in terms of "r" and "x". We can use similar triangles:
(r / R) = (h / (h - x))
Cross-multiplying and rearranging, we have:
r * (h - x) = R * h
h - x = (R * h) / r
h = (R * h) / r + x
We now substitute the expression for "h" in the volume ratio equation:
(3 * R^2 * x) / (r^2 * ((R * h) / r + x)) = 4/9
To simplify the equation further, let's substitute "H = R * h / r":
(3 * R^2 * x) / (r^2 * (H + x)) = 4/9
Now, we can solve for "x". Let's cross-multiply and rearrange the equation:
(27 * R^2 * x) = 4 * r^2 * (H + x)
Expanding both sides of the equation:
27 * R^2 * x = 4 * r^2 * H + 4 * r^2 * x
Isolating the "x" terms:
27 * R^2 * x - 4 * r^2 * x = 4 * r^2 * H
Factor out "x":
x * (27 * R^2 - 4 * r^2) = 4 * r^2 * H
Now, solve for "x" by dividing both sides of the equation:
x = (4 * r^2 * H) / (27 * R^2 - 4 * r^2)
This gives us the value of "x". To find the corresponding value of "R", we can substitute the expression for "x" back into the equation for "h":
h = (R * h) / r + x
h = H + x
We know that the height of the cone (h) is given, and we have already computed "H" in terms of "R" and "r". Now, we solve for "R":
H + x = (R * h) / r + x
H = (R * h) / r
R = (H * r) / h
Substituting the values of "H", "r", and "h" into the equation, we can find the value of "R" and "x". Then, we can use these values to calculate the volume of the cylinder using the formula:
Volume of the cylinder = π * R^2 * x
Finally, we substitute the values of "R" and "x" to find the maximum volume of the inscribed cylinder.