The volume of a cube of edge x is increasing at a rate of 12 cm/s. Find the rate of change of edge of the cube when the edge is 6 cm

The correct option is 1/9.

Let us denote the edge length of the cube by $x$ and its volume by $V$.
The volume $V$ of a cube with edge length $x$ is given by the formula:
$V=x^3$

We are given that the volume of the cube is increasing at a rate of $12{\text{cm}}^3\text{/s}$ (noting that rate of change of volume is with respect to time $t$). Therefore:
$\frac{dV}{dt}=12{\text{cm}}^3\text{/s}$

We need to find the rate of change of the edge, which is $\frac{dx}{dt}$, at the instant when the edge $x=6\text{cm}$.

By differentiating the volume formula with respect to time $t$ using the chain rule, we get:
$\frac{dV}{dt}=\frac{d}{dt}\left(x^3\right)=3x^2\cdot \frac{dx}{dt}$

Now, we substitute the known values into this equation:
$12=3x^2\cdot \frac{dx}{dt}$

Rearranging the equation to solve for $\frac{dx}{dt}$ gives:
$\frac{dx}{dt}=\frac{12}{3x^2}=\frac{4}{x^2}$

At the instant when $x=6\text{cm}$, we substitute this value into the expression for $\frac{dx}{dt}$:
$\frac{dx}{dt}=\frac{4}{6^2}=\frac{4}{36}=\frac{1}{9}\text{cm/s}$

Thus, the rate of change of the edge of the cube is $1/9\text{cm/s}$ when the edge length is 6 cm.