Answer is: option3
\( V \) is increasing only when \( r > h \).Solution:
\[ V = \frac{1}{3} \pi r^2 h \]
Using the product rule:\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} + h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]
- \(\frac{dr}{dt} = -2\) cm/min (since radius is decreasing)
- \(\frac{dh}{dt} = 4\) cm/min (since height is increasing)
\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2r(-2)h + r^2(4) \right) \]
\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( -4rh + 4r^2 \right) \]
\[ \frac{dV}{dt} = \frac{4}{3} \pi (r^2 - rh) \]
For volume to be increasing, we need:\[ r^2 - rh > 0 \]
\[ r(r - h) > 0 \]
This inequality holds when \( r > h \) or \( r < 0 \). Since radius cannot be negative, the condition simplifies to:
\[ r > h \]
Conclusion:- \( V \) is increasing only when \( r > h \).