Right cylinder from a height-radius relation

Text

In a right cylinder the height is $\dfrac32$ of the base radius. Knowing that $\dfrac79$ of the height plus $\dfrac56$ of the base diameter measures $34$ cm, find the total surface area and the volume.

Solution

Let $h=\dfrac32 r$. The condition is $\frac79\cdot\frac32 r+\frac56\cdot2r=\frac{7}{6}r+\frac{5}{3}r=\frac{7r+10r}{6}=\frac{17}{6}r=34\ \Rightarrow\ r=12,\ h=18.$ Total surface: $S_t=2\pi r(r+h)=2\pi\cdot12\cdot30=720\pi\approx2261.95$ cm$^2$. Volume: $V=\pi r^2 h=\pi\cdot144\cdot18=2592\pi\approx8143.01$ cm$^3$. $\boxed{S_t=720\pi\approx2261.95\ \text{cm}^2;\ V=2592\pi\approx8143.01\ \text{cm}^3}$