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 $6$ inches, what is the radius?
We are given the equation $2\pi r^2 + 2\pi rh = \frac{1}{3} \pi r^2 h$, but since $h=6$, $2\pi r^2 + 12\pi r = \frac{1}{3} \pi r^2 6$. Dividing by $\pi r$, we have $2r+12 = \frac{1}{3} r \cdot 6$. Solving this equation gives $r = \boxed{18}$.
To solve this problem, we can set up an equation using the given formulas and then solve for the radius.
Given that the height of both the cylinder and the cone is 6 inches, we know that $h = 6$.
Let's plug in the values into the formulas:
For the surface area of the cylinder: $SA_{cylinder} = 2\pi r^2 + 2\pi rh$
For the volume of the cone: $V_{cone} = \frac{1}{3} \pi r^2 h$
But, since the surface area is numerically equal to the volume of the cone, we can set the two formulas equal to each other:
$2\pi r^2 + 2\pi rh = \frac{1}{3} \pi r^2 h$
Now let's substitute $h = 6$ into the equation:
$2\pi r^2 + 2\pi r(6) = \frac{1}{3} \pi r^2 (6)$
Simplifying the equation further:
$2\pi r^2 + 12\pi r = 2\pi r^2$
Notice that both sides of the equation have $\pi r^2$. We can now cancel those terms:
$12\pi r = 0$
Since $\pi$ is a non-zero constant, we can divide by $\pi$ on both sides of the equation:
$12r = 0$
Dividing by 12:
$r = 0$
However, a radius cannot be zero in this context. Therefore, there is no valid radius for which the surface area of the cylinder is numerically equal to the volume of the cone when the height is 6 inches.