It is given in the question that
The revenue (in thousands of dollars) from producing x units of an item is modeled by
![R(x)=12x-0.01x^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7f69o8psl1zo9eqpyb4ms2qsp0dgvmkz1l.png)
a. Average rate of change is given by
![(R(1007)-R(1002))/(1007-1002)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gyrze7329cbnoj0rirqo2tvl4f38ulyqya.png)
![= (1943.51-1983.96)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jbv1o268togzsuhwwmq2mf15vweq6tipmr.png)
![=(-40.45)/(5) = -8.09](https://img.qammunity.org/2019/formulas/mathematics/middle-school/oinkdy63ndamvruastk4n9htzjgeip55qm.png)
So the average rate of change of revenue is -8090dollars per unit .
b .To find the marginal revenue, we have to differentiate revenue function, that is
![R'(x)=12-0.02x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/f8oop4o08j51nej6hwhqzz8eehyljh0vmk.png)
And at x=1000, we will get
![R'(1000) = 12-0.02(1000)=-8](https://img.qammunity.org/2019/formulas/mathematics/middle-school/uvutu9moa5uvnnd765g2hr7lelv6ug57uf.png)
S the marginal revenue is -8000 dollars.