SOLUTION 5: Draw a cube with edge lengths $x$, and assume that $x$ is a function of time $t$.

$\ \ \ \$ a.) The surface area (Add the areas of 6 square surfaces.) of a cube is $$S=x^2+x^2+x^2+x^2+x^2+x^2 \ \ \ \ \longrightarrow$$ $$S=6x^2$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$

FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $x =80 \ cm.$

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

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

$\Big($ Now let $\displaystyle{ dx \over dt } = -2$ and $x=80. \Big)$ $$\displaystyle{ dS \over dt } = 12(80)(-2) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = -1920 \ cm^2/min.$$

$\ \ \ \$ b.) The volume of a cube is $$V = (length)(width)(height) \ \ \ \ \longrightarrow$$ $$V = x^3$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$

FIND: $\ \ \ \displaystyle{ dV \over dt }$ when $x =80 \ cm.$

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 $\displaystyle{ dx \over dt } = -2$ and $x=80. \Big)$

$$\displaystyle{ dV \over dt } = 3(80)^2(-2) = -38,400 \ \ cm^3/min.$$