The surface area of a solid cylinder is found using the formula $SA = 2\pi r^2 + 2\pi rh$. The volume of a cone is found using the formula $V = \frac{1}{3} \pi r^2 h$. Judy has a solid cylinder whose total surface area is numerically equal to the volume of a particular cone that has a base congruent to the cylinder's base. If the height of both solids is $4$ inches, what is the radius?
Setting the surface area of the cylinder equal to the volume of the cone, we have \[2\pi r^2 + 2\pi rh = \frac13 \pi r^2 h.\]We divide by $\pi r$: \[2r + 2h = \frac13 r h.\](That is, we divide by $\pi r^2$ and multiply by $3$ to cancel the factor of $\frac13 \pi r^2 h$ on the right side.) Substituting $h = 4,$ we have \[2r + 2(4) = \frac13 r(4).\]Simplifying the equation gives $2r + 8 = \frac{4}{3}r,$ so $\frac{2}{3}r = 8,$ or $r = \boxed{12}.$
To solve this problem, we need to equate the surface area of the solid cylinder to the volume of the cone.
Given:
Height of both solids, $h = 4$ inches
Let's simplify the expressions for the surface area and volume using the given height:
Surface area of solid cylinder:
$SA_{cylinder} = 2\pi r^2 + 2\pi rh$
Volume of cone:
$V_{cone} = \frac{1}{3}\pi r^2 h$
Since the base of the cone is congruent to the base of the cylinder, we can equate the two:
$SA_{cylinder} = V_{cone}$
$2\pi r^2 + 2\pi rh = \frac{1}{3}\pi r^2 h$
Now, substitute the given height $h=4$ into the equation:
$2\pi r^2 + 2\pi r(4) = \frac{1}{3}\pi r^2 (4)$
Simplify the equation:
$2\pi r^2 + 8\pi r = \frac{4}{3}\pi r^2$
Multiply both sides by $3$ to eliminate the denominator:
$6\pi r^2 + 24\pi r = 4\pi r^2$
Rearrange the equation:
$4\pi r^2 - 6\pi r^2 + 24\pi r = 0$
Simplify and factor out $\pi r$:
$-2\pi r(r - 12) = 0$
We have two possible solutions:
1. $-2\pi r = 0$
2. $r - 12 = 0$
Let's consider the second solution:
$r - 12 = 0$
Adding $12$ to both sides, we find:
$r = 12$
Therefore, the radius of the solid cylinder is $12$ inches.