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A cylinder is placed inside a cube

This topic has 4 expert replies and 1 member reply

A cylinder is placed inside a cube

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A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described?

A. 16/π
B. 2π
C. 8
D. 4π
E. 8π

Can some experts help me find the best solution in this problem?

OA D

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lheiannie07 wrote:
A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described?

A. 16/π
B. 2π
C. 8
D. 4π
E. 8π
The cylinder of the maximum volume that can be inscribed in a cube is one with the diameter of its base being the side length of the cube and the height also being the side length of the cube. Recall that the volume of a cylinder is V = πr^2h, where r is the radius and h is the height. If s = side length of the cube, we have r = s/2 (since s is also the length of the diameter) and h = s. Thus, the maximum volume of the cylinder is:

V = π*(s/2)^2*(s)

V = π(s^3)/4

Notice that s^3 is the volume of the cube and it’s given to be 16; thus, the maximum volume of the cylinder is:

V = π(16)/4

V = 4π

Answer: D

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lheiannie07 wrote:
A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described?

A. 16/π
B. 2π
C. 8
D. 4π
E. 8π

Can some experts help me find the best solution in this problem?

OA D
If 2a is the side of the cube its volume would be 8a^3=16
a^3= 2
Diameter and height of the biggest cylinder that fits in cube would be 2a
and its volume would be =π(a)^2X2a=2πa^3 =4 π…………( because a^3=2)
hence option D is correct.
.

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If the volume of the cube is 16, each of its sides are $$\sqrt[3]{16}$$ . The largest possible cylinder will then have a diameter AND height of $$\sqrt[3]{16}$$. We find the area of a cylinder with $$V\ =\ \pi r^2\cdot h$$ Plugging in values (remembering that r = d/2) gives $$V\ =\ \pi\left(\frac{\sqrt[3]{16}}{2}\right)^2\cdot\sqrt[3]{16}$$ $$V\ =\ \pi\frac{\sqrt[3]{16}}{2}\cdot\frac{\sqrt[3]{16}}{2}\cdot\sqrt[3]{16}$$ $$V\ =\ \pi\frac{16}{4}$$ $$V\ =\ 4\pi$$

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lheiannie07 wrote:
A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described?

A. 16/π
B. 2π
C. 8
D. 4π
E. 8π
The volume of the cube is 16
Volume of cube = (side length)³
So: 16 = (side length)³
So, side length = ∛16

So, the BASE of the cube is a SQUARE with dimension ∛16 by ∛16
So, the largest cylinder to fit inside the cube must have a diameter of ∛16
This means the RADIUS of the cylinder = ∛16/2

Also, since the cube has HEIGHT ∛16, the largest cylinder to fit inside the cube must have a HEIGHT of ∛16

What is the maximum possible volume of the cylinder that fits inside the cube as described?
Volume = π(radius)²(height)
= π(∛16/2)²(∛16)
= π(16/4)
= 4π
= D

Cheers,
Brent

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Quote:
A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described?

A. 16/π
B. 2π
C. 8
D. 4π
E. 8π
Hi lheiannie07,
Let's take a look at your question.

Volume of the cube = 16
If x represents the side length of the cube then,
$$x^3=16$$
$$x=\sqrt[3]{16}$$

Since, the cylinder is placed inside the cube, therefore its maximum radius and height will be equal to the side length of the cube.
Hence,
$$Height=h=\sqrt[3]{16}$$
$$Diameter=\sqrt[3]{16}$$
$$Radius=r=\frac{\sqrt[3]{16}}{2}$$

Volume of the cylinder canbe calculated using formula:
$$=\pi r^2h$$
$$=\pi\left(\frac{\sqrt[3]{16}}{2}\right)^{^{^2}}\left(\sqrt[3]{16}\right)$$
$$=\pi\left(\frac{16^{\frac{2}{3}}}{4}\right)\left(16^{\frac{1}{3}}\right)$$
$$=\frac{\pi}{4}.\left(16^{\frac{2}{3}+\frac{1}{3}}\right)$$
$$=\frac{\pi}{4}.\left(16^{\frac{3}{3}}\right)$$
$$=\frac{\pi}{4}.\left(16\right)$$
$$=4\pi$$

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.

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