Solve for the indicated letter.
-Solve the formula $S=\frac{1}{3} \pi r^{2} h+4 \pi r h$ for $r$.
A) $r=\frac{12 \pi h \pm \sqrt{36 \pi^{2} h^{2}+3 \pi h S}}{\pi h}$
B) $r=\frac{-6 \pi h \pm \sqrt{36 \pi^{2} h^{2}+3 \pi h S}}{\pi h}$
C) $r=\frac{-6 \pi h \pm 2 \sqrt{36 \pi^{2} h^{2}+3 \pi h S}}{\pi h}$
D) $r=\frac{-12 \pi \mathrm{h} \pm \sqrt{36 \pi^{2} \mathrm{~h}^{2}+3 \pi \mathrm{hS}}}{2 \pi \mathrm{h}}$

Question

Quizplus · Accepted Answer

To solve for $r$, we start with the given equation $S=\frac{1}{3} \pi r^{2} h+4 \pi r h$. First, we factor out $ \pi h $ to get $S = \pi h (\frac{1}{3}r^2 + 4r)$. Then, we rearrange the equation to form a quadratic equation in terms of $r$: $\frac{1}{3}r^2 + 4r - \frac{S}{\pi h} = 0$. Multiplying through by 3 to clear the fraction gives $r^2 + 12r - \frac{3S}{\pi h} = 0$. Applying the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=12$, and $c=-\frac{3S}{\pi h}$, we get $r = \frac{-12 \pm \sqrt{144 + 12S/\pi h}}{2}$. Simplifying further and correcting for the setup leads to $r=\frac{-6 \pi h \pm \sqrt{36 \pi^{2} h^{2}+3 \pi h S}}{\pi h}$, which matches option B.

Solve for the Indicated Letter S=13πr2h+4πrhS=\frac{1}{3} \pi r^{2} h+4 \pi r hS=31​πr2h+4πrh

Solve for the Indicated Letter $S=\frac{1}{3} \pi r^{2} h+4 \pi r h$