Question

A toy company is considering a cube or sphere-shaped container for packaging a new product. The height of the cube would equal the diameter of the sphere. Compare the volume-to-surface area ratios of the containers. Which packaging will be more efficient?

Answer 1

For a sphere \[SA = 4 \pi r ^{2}\]

Answer 2

ok... so the volume of the cube is

\[V = x^3 \]

for the shpere if the diameter is x then the radius is x/2

the volume of the sphere is

\[V = \frac{4}{3} \pi r^3\]

substitute r = x/2 and evaluate

Answer 3

surface area of the cube is

\[SA = 6x^2\]

for the surface are of the sphere its

\[SA = 4 \pi r^2\]

substitute r = x/2

hope this helps..

Answer 4

@campbell_st so would I use both the surface area formulas and the volume formulas for this ?

Answer 5

they are asking you to find a ratio Volume to Surface area so to me I do

Cube

\[\frac{V}{SA} = \frac{x^3}{6x^2} = \frac{x}{6}\]

you would need a similar relationship for the sphere...

then I'd assume the smaller number would be more efficient...

Answer 6

okay so would the sphere one look like \[\frac{ V }{ SA } = \frac{ 4/3 }{ 4 } = 3 ?\]

Answer 7

Im kind of confused on how you would do the pi and radius sign in here

Answer 8

yep... so volume of the sphere is

\[V = \frac{4}{3} \pi (\frac{x}{2})^3 = \frac{\pi x^3}{6}\]

and

\[SA = \pi x^2\]

the interesting thing is the ratios are equal...

Sphere..

\[\frac{V}{SA} = \frac{\frac{\pi x^3}{6}}{\pi x^2}\]

Answer 9

Ohh so the answer would be that they are both equally efficient?

Answer 10

well thats what the math says... providing there are correct...

the ratio for the Sphere is

\[\frac{V}{SA} = \frac{x}{6}\]

Answer 11

oops should be providing the math is correct...

Answer 12

Ohh okay! Well thank you so much for your help..again lol

Answer 13

just hope it didn't confuse... good luck

Answer 14

No, I understand, a lot better than before actually so thank you