Answer
(a). The rate at which the volume is changing: 6 $m^{3}/s$
(b). The rate at which the surface area is changing: 10 $m^{2}/s$
(c). The rate at which the length of a diagonal is changing: 0 $m/s$
Work Step by Step
(a). The volume is $V=lwh$ , use the Chain rule, we get
$$\frac{dV}{dt}=\frac{\partial{V}}{\partial{l}}\frac{dl}{dt}+\frac{\partial{V}}{\partial{w}}\frac{dw}{dt}+\frac{\partial{V}}{\partial{h}}\frac{dh}{dt}=(hw)\frac{dl}{dt}+(lh)\frac{dw}{dt}+(lw)\frac{dh}{dt}$$
$$\Rightarrow\frac{dV}{dt}=(2\times2)\times2+(1\times2)\times2+(1\times2)\times(-3)=6m^{3}/s$$
(b). The surface area is $S=2lw+2lh+2hw$ , by the Chain rule, we get
$$\frac{dS}{dt}=\frac{\partial{S}}{\partial{l}}\frac{dl}{dt}+\frac{\partial{S}}{\partial{w}}\frac{dw}{dt}+\frac{\partial{S}}{\partial{h}}\frac{dh}{dt}=2(w+h)\frac{dl}{dt}+2(l+h)\frac{dw}{dt}+2(l+w)\frac{dh}{dt}$$
$$\Rightarrow\frac{dS}{dt}=2(2+2)\times2+2(1+2)\times2+2(1+2)\times(-3)=10m^{2}/s$$
(c). We know $L^{2}=l^{2}+w^{2}+h^{2}$ , differentiate both sides with respect to t, we get
$$2L\frac{dL}{dt}=2l\frac{dl}{dt}+2w\frac{dw}{dt}+2h\frac{dh}{dt}$$
$$\Rightarrow2L\frac{dL}{dt}=2\times1\times2+2\times2\times2+2\times2\times(-3)=0$$
Since $L\ne0$ , thus $\frac{dL}{dt}=0 m/s$.
