SOLUTION 12: Draw a cube with edge length $x$, and assume that $x$ is a function of time $t$.

The surface area (Add the areas of 6 square surfaces.) of a cube is

$$S = x^2+x^2+x^2+x^2+x^2+x^2 \ \ \ \ \longrightarrow$$ $$S=6x^2$$

The volume of a cube is $$V = (length)(width)(height) \ \ \ \ \longrightarrow$$ $$V = x^3$$

GIVEN: $\ \ \ \displaystyle{ dS \over dt } = 600 \ in^2/hr.$

FIND: $\ \ \ \displaystyle{ dV \over dt }$ when $x =10 \ in.$

Now differentiate the surface area equation with respect to time $t$ getting

$$D \{ S \} = D \{ 6x^2 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 6 \cdot 2x \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 12 x \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dS \over dt } = 600$ and $x=10. \Big)$

$$600 = 12(10) \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$ $$\displaystyle{ dx \over dt } = 5 \ in/hr.$$

Now differentiate the volume equation with respect to time $t$ getting

$$D \{ V \} = D \{ x^3 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = 3 x^2 \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 5$ and $x=10. \Big)$

$$\displaystyle{ dV \over dt } = 3(10)^2(5) = 1500 \ \ in^3/hr.$$