The total accumulated costs $C(t)$ and revenues $R(t)$ (in thousands of dollars), respectively, for a photocopying machine satisfy \[ C^{\prime}(t)=\frac{1}{21} t^{7} \quad \text { and } \quad R^{\prime}(t)=7 t^{7} e^{-t^{8}} \] where $t$ is the time in years. Find the useful life of the machine, to the nearest year. What is the total profit accumulated during the useful life of the machine? The useful life of the machine is $\square$ year(s). (Round to the nearest year as needed.) Using the useful life of the machine rounded to the nearest year, the total profit accumulated during the useful life of the machine is $\$ \square$. (Round to the nearest dollar as needed.)

Step 1 ：Given that the derivative of the revenue function is \(R^{\prime}(t)=7 t^{7} e^{-t^{8}}\), we set this equal to zero and solve for \(t\).

Step 2 ：Solving \(7 t^{7} e^{-t^{8}} = 0\), we find that this equation is satisfied when \(t=0\) or when \(e^{-t^{8}}=0\). However, the exponential function is never zero, so the only critical point is \(t=0\).

Step 3 ：To determine whether this is a maximum or minimum, we need to consider the second derivative of the revenue function. The second derivative is given by \(R^{\prime\prime}(t) = 7 t^{6} e^{-t^{8}} (8t - 7)\).

Step 4 ：Evaluating this at \(t=0\) gives \(R^{\prime\prime}(0) = 0\), which does not provide any information about the nature of the critical point. However, we know that the revenue function is increasing for \(t>0\) (since the first derivative is positive for \(t>0\)) and decreasing for \(t<0\) (since the first derivative is negative for \(t<0\)). Therefore, the critical point at \(t=0\) is a maximum.

Step 5 ：So, the useful life of the machine is \(\boxed{0}\) years.

Step 6 ：The total cost is given by \(C(t) = \int C^{\prime}(t) dt = \int \frac{1}{21} t^{7} dt = \frac{1}{168} t^{8} + C\), where \(C\) is the constant of integration. Evaluating this at \(t=0\) gives \(C(0) = C = 0\), so the total cost is \(C(t) = \frac{1}{168} t^{8}\).

Step 7 ：The total revenue is given by \(R(t) = \int R^{\prime}(t) dt = \int 7 t^{7} e^{-t^{8}} dt\). This integral is not elementary, so we cannot find an explicit expression for \(R(t)\). However, we know that the total revenue is maximized at \(t=0\), so the total revenue at the useful life of the machine is \(R(0) = 0\).

Step 8 ：Therefore, the total profit accumulated during the useful life of the machine is \(R(0) - C(0) = 0 - 0 = 0\).

Step 9 ：So, the total profit accumulated during the useful life of the machine is \(\boxed{0}\).