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

Answer:

Packaging with cube will be more efficient.

Step-by-step 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 =  $(\frac{x}{2})$

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 = $\frac{4\pi r^3}{3}$

We have to compare the ratio of SA and Volumes.

Ratio of SA :                                        Ratio of their volumes :

⇒ $\frac{SA\ of\ cube\ (S_1)}{SA\ of\ sphere\ (S_2)}$                            ⇒ $\frac{Volume \ of \ cube\ (V_1)}{Volume\ of\ sphere\ (V_2)}$

⇒ $\frac{6x^2}{4\pi (\frac{x}{2})^2}$                                           ⇒ $\frac{x^3}{\frac{4 \pi r^3}{3} }$

⇒ $\frac{6x^2}{4\pi (\frac{x^2}{4})}$                                           ⇒ $\frac{x^3}{\frac{4 \pi (\frac{x}{2})^3}{3} }$

⇒ $\frac{6x^2}{\pi x^2}$                                               ⇒  $\frac{x^3}{\frac{4 \pi (\frac{x^3}{8})}{3} }$

⇒ $\frac{6}{\pi}$                                                  ⇒ $\frac{6}{\pi}$

⇒ 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.