We have to find the marginal cost of the function:
[tex]C(t)=140-30e^{-t}[/tex]
for the value t=6. We remember that the marginal cost is defined as the derivative of the function of total cost. So, for finding the value of marginal cost, we will find its function:
[tex]M(t)=C^{\prime}(t)=\frac{d}{\differentialDt t}(140-30e^{-t})[/tex]
Then, using the properties of differentiation,
[tex]\begin{gathered} \frac{d}{\differentialDt t}(140-30e^{-t}) \\ =\frac{d}{\differentialDt t}(140)-\frac{d}{\differentialDt t}(30e^{-t^{}}) \\ =0-30\frac{d}{\differentialDt t}(e^{-t}) \\ =-30\frac{d}{\differentialDt t}(e^{-t}) \end{gathered}[/tex]
And then, for finding the last derivative, we use the chain rule:
[tex]=-30(-1)e^{-t}=30e^{-t}[/tex]
This means that our marginal cost function is:
[tex]M(t)=30e^{-t}[/tex]
Finding the value when t=6
We just have to find the value M(6), by replacing t by 6, as shown:
[tex]M(6)=30e^{-6^{}}=\frac{30}{e^6}=0.0743625653\approx0.07[/tex]
This means that the marginal cost when t=6 is 0.07 million dollars per year.