Let the side of the cube be $a$ cm.

Then, $\text{Surface area } S = 6a^2$ $\text{Volume } V = a^3$

Differentiate surface area with respect to time $t$.

Given that the surface area is increasing at the rate $\frac{dS}{dt} = 3.6 \text{ cm}^2/\text{sec}$

Now, $S = 6a^2$ Differentiating w.r.t. $t$, $\frac{dS}{dt} = 12a\frac{da}{dt}$

Substitute $a=10$ and $\frac{dS}{dt}=3.6$: $3.6 = 12(10)\frac{da}{dt}$ $3.6 = 120\frac{da}{dt}$ $\frac{da}{dt} = \frac{3.6}{120} = 0.03 \text{ cm/sec}$

Differentiate volume with respect to time.

$V = a^3$ So, $\frac{dV}{dt} = 3a^2\frac{da}{dt}$

At $a=10$ and $\frac{da}{dt}=0.03$, $\frac{dV}{dt} = 3(10)^2(0.03)$ $\frac{dV}{dt} = 3(100)(0.03) = 9$

Final answer

The rate of change of volume is $\boxed{9 \text{ cm}^3/\text{sec}}$

Option check

• A: $9$ ✅
• B: $10$ ❌
• C: $18$ ❌
• D: $20$ ❌

Comparison with stored answer

Stored correct answer is A, which matches the derived answer.