The radius of a right circular cylinder is increasing at the rate of…

The correct answer is Option (1) → an increase of \(48\pi\) cm³/s

Let r be the radius and h the height at time t, with \(\frac{dr}{dt} = 2\) and \(\frac{dh}{dt} = -3\) (the height is decreasing). The volume is

    \( V = \pi r^{2}h \)

Both r and h change, so by the product rule:

    \( \displaystyle \frac{dV}{dt} = \pi\left(2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt}\right) \)

With r = 4 and h = 6:

    \( \displaystyle \frac{dV}{dt} = \pi\left(2 \times 4 \times 6 \times 2 + 16 \times (-3)\right) \)

    \( = \pi(96 - 48) = 48\pi \)

The value is positive, so the volume is increasing at \(48\pi\) cm³/s.

Why the other options are incorrect

• Option (2) an increase of \(144\pi\) cm³/s — Incorrect. This is \(\pi(96 + 48)\): the height term is added although the height is decreasing.

• Option (3) a decrease of \(48\pi\) cm³/s — Incorrect. The size is right, but \(\frac{dV}{dt} = +48\pi \gt 0\), so the volume increases.

• Option (4) an increase of \(96\pi\) cm³/s — Incorrect. This keeps only the radius term \(2\pi rh\frac{dr}{dt}\) and leaves out the change of height.