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