1. Monomials

A ball with radius r fits snugly in a cubical box. Determine the ratio of the surface area of the box to the surface area of the ball.
The surface area of a box is 6s^2 and the surface area of a sphere is 4pi(r^2)
The answer is supposed to be 6/pi
I think that would be
6(r^2) over 4pi(r^2)
the two (r^2) cancel out so I'm left with 6/4pi
I don't know what to do from there.

help!

2. Since s=2r, make the sub.

$\displaystyle \frac{6(2r)^{2}}{4{\pi}r^{2}}=\frac{24r^{2}}{4{\pi }r^{2}}=\frac{6}{\pi}$

3. oh! thanks

4. We know the surface area of the box(assuming that it is a cube) is given by $\displaystyle 6l^2$, where $\displaystyle l$ is the length of one of the sides of the cube.

We also know the equation for the surface area of a sphere is $\displaystyle 4\pi r^2$, where $\displaystyle r$ is the radius of the sphere.

In this scenario, the radius of the sphere would be half of the length of one of the sides of the cube, as it fits snuggly inside the box. So we have that:

$\displaystyle r = \frac {l}{2}$

We then have that the surface area of the ball is:

$\displaystyle 4\pi (\frac {l}{2})^2$

Then we simply divide this expression into the expression representing the area of the box:

$\displaystyle \frac {6l^2}{4\pi (\frac {l}{2})^2}$

Simplify to get:

$\displaystyle \frac {6l^2}{\frac {4\pi l^2}{4}}$

We then have a final answer of:

$\displaystyle \frac {6}{\pi}$