Step 1

Let the surface area of the cube be

$ S=6x^2$

and the volume of the cube be

$v=x^3$

where $x$ is the edge of the cube

$v=x^3$

differentiating w.r.t $x$ we get,

$ \large\frac{dv}{dt}=3x^2\large\frac{dx}{dt}$

But it is given that $ \large\frac{dv}{dt}=k $ ( constant)

$ \therefore k = 3x^2\large\frac{dx}{dt} \Rightarrow \large\frac{dx}{dt}=\large\frac{k}{3x^2}$

$ s=6x^2$

differentiating w.r.t $x$ we get,

$ \large\frac{ds}{dt}=12x.\large\frac{dx}{dt}$

Now substituting for $\large\frac{dx}{dt}$ we get,

$\large\frac{ds}{dt}=\large\frac{k}{3x^2} \times 12x$

$=\large\frac{4k}{x}$

$ \Rightarrow \large\frac{ds}{dt} \alpha \large\frac{1}{x}$

Hence the rate of increase in surface area varies inversely as the length of the edge of the cube.