SOLUTION 10: Consider the given closed rectangular box with dimensions $x, y,$ and $z$, and assume that $x, y,$ and $z$ are functions of time $t$.

$\ \ \ \$ a.) The surface area of the box (Add the areas of six rectangular surfaces.) is $$S= xy+xy+xz+xz+yz+yz \ \ \ \ \longrightarrow$$ $$S= 2xy+2xz+2yz$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/hr.$, $\displaystyle{ dy \over dt } = -6 \ ft/hr.$, and $\displaystyle{ dz \over dt } = 3 \ ft/hr.$

FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $x=5 \ ft.$, $y=4 \ ft.$, and $z=6 \ ft.$

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

$$D \{ S \} = D \{ 2xy + 2xz + 2yz \} \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 2 ( x \displaystyle{ dy \over dt } + \displaystyle{ dx \over dt } y ) + 2 ( x \displaystyle{ dz \over dt } + \displaystyle{ dx \over dt } z ) + 2 ( y \displaystyle{ dz \over dt } + \displaystyle{ dy \over dt } z ) \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4, \displaystyle{ dy \over dt } = -6, \displaystyle{ dz \over dt } = 3, x=5, y=4$, and $z=6. \Big)$

$$\displaystyle{ dS \over dt } = 2 ( (5)(-6) + (4)(4) ) + 2 ( (5)(3) + (4) (6) ) + 2 ( (4)(3) + (-6) (6) ) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 2 ( -14 ) + 2 ( 39 ) + 2 ( -24 ) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = -28 + 78 + -48 \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 2 \ ft^2/min.$$

$\ \ \ \$ b.) The volume of the box is $$V= (length)(width)(height) \ \ \ \ \longrightarrow$$ $$V=xyz$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/hr.$, $\displaystyle{ dy \over dt } = -6 \ ft/hr.$, and $\displaystyle{ dz \over dt } = 3 \ ft/hr.$

FIND: $\ \ \ \displaystyle{ dV \over dt }$ when $x=5 \ ft.$, $y=4 \ ft.$, and $z=6 \ ft.$

Now differentiate the volume equation with respect to time $t$ using the triple product rule, getting

$$D \{ V \} = D \{ xyz \} \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = \displaystyle{ dx \over dt }yz + x \displaystyle{ dy \over dt }z + xy\displaystyle{ dz \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4, \displaystyle{ dy \over dt } = -6, \displaystyle{ dz \over dt } = 3, x=5, y=4$, and $z=6. \Big)$

$$\displaystyle{ dV \over dt } = (4)(4)(6) + (5)(-6)(6) + (5)(4)(3) \ \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = 96-180+60 \ \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = -24 \ ft^3/hr.$$

$\ \ \ \$ c.) Consider the given closed rectangular box with dimensions $x$ feet by $y$ feet by $z$ feet, and the main diagonal of length $L$ feet in the given diagram.

Consider the two right triangles of edge lengths $x, y,$ and $w$ and $w, z,$ and $L$, resp. Apply the Pythagorean Theorem to each right triangle getting $$x^2+y^2=w^2 \ \ \ and \ \ \ w^2 + z^2 = L^2$$

Substituting for $w^2$ leads to the following main diagonal equation:

$$L^2 = x^2+y^2 + z^2$$

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/hr.$, $\displaystyle{ dy \over dt } = -6 \ ft/hr.$, and $\displaystyle{ dz \over dt } = 3 \ ft/hr.$

FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $x=5 \ ft.$, $y=4 \ ft.$, and $z=6 \ ft.$

Now differentiate the main diagonal equation with respect to time $t$, getting

$$D \{ L^2 \} = D \{ x^2+y^2 + z^2 \} \ \ \ \longrightarrow$$ $$2L \displaystyle{ dL \over dt } = 2x \displaystyle{ dx \over dt } + 2y \displaystyle{ dy \over dt } + 2z \displaystyle{ dz \over dt } \ \ \ \longrightarrow$$ $$L \displaystyle{ dL \over dt } = x \displaystyle{ dx \over dt } + y \displaystyle{ dy \over dt } + z \displaystyle{ dz \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4, \displaystyle{ dy \over dt } = -6, \displaystyle{ dz \over dt } = 3, x=5, y=4$, and $z=6 .$ The equation $L^2 = x^2+y^2+z^2$ becomes $L^2=5^2+4^2+6^2$, so that $L^2=77$ and $L=\sqrt{77}. \Big)$

$$\sqrt{77} \displaystyle{ dL \over dt } = (5)(4) + (4)(-6) + (6) (3) \ \ \ \ \longrightarrow$$ $$\sqrt{77} \displaystyle{ dL \over dt } = 14 \ \ \ \ \longrightarrow$$ $$\displaystyle{ dL \over dt } = { 14 \over \sqrt{77} } \ ft/hr. \approx 1.595 ft/hr.$$