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? For a sphere, SA =4πr²

Answer 1

Packaging with cube will be more efficient.

Explanation:

Given:

A cube and a sphere where the diameter of the sphere is equal to the height of the cube.

Let the height of the cube "x"

Radius of the sphere =
`((x)/(2))`

Formula to be used:

Surface area of the cube =
`6x^2` and Surface area of the sphere =
`4\pi (r)^2`

Volume of the cube =
`x^3` and Volume of the sphere =
`(4\pi r^3)/(3)`

We have to compare the ratio of SA and Volumes.

Ratio of SA : Ratio of their volumes :

⇒
`(SA\ of\ cube\ (S_1))/(SA\ of\ sphere\ (S_2))` ⇒
`(Volume \ of \ cube\ (V_1))/(Volume\ of\ sphere\ (V_2))`

⇒
`(6x^2)/(4\pi ((x)/(2))^2)` ⇒
`(x^3)/((4 \pi r^3)/(3) )`

⇒
`(6x^2)/(4\pi ((x^2)/(4)))` ⇒
`(x^3)/((4 \pi ((x)/(2))^3)/(3) )`

⇒
`(6x^2)/(\pi x^2)` ⇒
`(x^3)/((4 \pi ((x^3)/(8)))/(3) )`

⇒
`(6)/(\pi)` ⇒
`(6)/(\pi)`

⇒ approx
`2` ⇒ approx
`2`

⇒
`S_1=2S_2` ⇒
`V_1=2V_2`

Packaging of the toy with the cube will be more efficient as it has more volume comparatively.