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QUESTION The time to failure,t in hours, of components follows the
  density function $f(t)=\alpha ^2t e{-\alpha t}, t>0$.\\

   \begin{description}

    \item[(a)]
     What is the probability that a component which has survived
     for two hours will fail in the next hour?

    \item[(b)]
     The cost of producing a component is $2\mu ^2$ units where
     $\mu$ is the mean time for failure. An income of 48t units
     per component is received for the time the component is
     working properly. Show that the maximum expected profit per
     component is 288 units corresponding to $\alpha=\frac{1}{6}$.
   \end{description}

ANSWER
 $f(t)=\alpha^2te^{-\alpha t},\ t>o$\\

  \begin{eqnarray*}
   F(t)&=&\int_0^t \alpha^2se^{-\alpha s}\,ds\\
   &=&[-\alpha s e^{-\alpha} s]_o^t+\int_0^t\alpha e^{-\alpha s}\,ds\\
   &=&-\alpha te^{-\alpha t}-[e^{-\alpha s}]_0^t\\
   &=&1-e^{-\alpha t}(1+\alpha t)
  \end{eqnarray*}

  \begin{description}

   \item[(a)]

    \begin{eqnarray*}
     \textrm{Require} P(2,T,3|t>2)&=&\frac{P(2<T<3 \textrm{ and
     }T>2}{P(T>2)}\\
     &=&\frac{P(2<T<3)}{P(T>2)}=\frac{F(3)-F(2)}{1-F(2)}\\
     &=&\frac{1-e^{-3\alpha}(1+3\alpha
     )-(1-e^{-2\alpha}(1+2\alpha))}{e^{-2\alpha}(1+2\alpha)}\\
     &=&{e^{-2\alpha}(1+2\alpha)-e^{-3\alpha}(1+3\alpha)}{e^{-2\alpha}(1+2\alpha)}\\
     &=&1+\frac{e^{-\alpha}(1+3\alpha)}{1+2\alpha}
    \end{eqnarray*}

   \item[(b)]

    \begin{eqnarray*}
     \mu&=&\int_0^\infty \alpha^2t^2e^{_\alpha t}\,dt\\
     &=&[-\alpha e^{-\alpha
     t}t^2]_0^\infty+\int_0^\infty2t\alpha e^{-\alpha t}\,dt\\
     &=&0+\frac{2}{\alpha}\int_0^\infty\alpha^2te^{-\alpha t}\,dt\\
     &=&\frac{2}{\alpha}\ \ \ (\textrm{since} \int f(t)\,dt=1)
    \end{eqnarray*}

    Profit$=48T-2\mu^2$\\
    E(Profit)$=48E(T)-\mu^2=48\mu-2\mu^2$\\
    $\frac{dE(\textrm{Profit})}{d\mu}=48-4\mu=0$ when $\mu=12$\\
    $\frac{d^2E(\textrm{Profit})}{d\mu^2}=-4$ therefore profit is
    maximum when $\mu=12$. Hence $\frac{2}{\alpha}=12$ therefore
    $\alpha=\frac{1}{6}$

    \end{description}


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