Chapter 4 - Calculating the Derivative - 4.1 Techniques for Finding Derivatives - 4.1 Exercises - Page 208: 53

$990$

We are given $p=\frac{1000}{q^2} + 1000$ From the given function for $p$, the revenue function is given by $R(q)=qp$ $=q(\frac{1000}{q^2} + 1000)$ $=\frac{1000}{q} + 1000q$ The marginal revenue is: $R'(q)=\frac{-1000}{q^2}+1000$ When $q=10$ $R'(q)=\frac{-1000}{(10)^2}+1000=990$