Given a fixed surface area for a cube and a sphere.
A= 1m^2
Determine the volume of the cube and the sphere. Which one has greater volume?
Well...the surface area of a square is $\displaystyle SA=6s^2$, where s is the side of the cube, and $\displaystyle SA=4\pi r$, where r is the radius of the sphere.
First find s and r, given that SA=1, then find the volume, noting that $\displaystyle V_{\text{cube}}=s^3$, and $\displaystyle V_{\text{sphere}}=\tfrac{4}{3}\pi r^3$.
Do these processes make sense?
--Chris
Yes, where SA=1.
So you would be solving $\displaystyle 1=6s^2$ for s, and $\displaystyle 1=4\pi r^{\color{red}2}$ for r.
Then you would plug these values into the corresponding volume equations I gave you. Then see which one is larger.
--Chris
EDIT: I caught a tiny mistake on my part. The surface area of a sphere is given by $\displaystyle SA=4\pi r^{\color{red}2}$, not $\displaystyle SA=4\pi r$.
My mistake!
How did you get those values?
$\displaystyle 1=6s^2\implies \tfrac{1}{6}=s^2\implies s=\sqrt{\tfrac{1}{6}}\approx \color{red}0.41$
$\displaystyle 1=4\pi r^2\implies r^2=\frac{1}{4\pi}\implies r=\sqrt{\frac{1}{4\pi}}\approx \color{red}0.28$.
Now that we know what s and r are, find $\displaystyle V_{\text{cube}}$ and $\displaystyle V_{\text{sphere}}$.
--Chris