## Calculus with Applications (10th Edition)

We are given $p=\frac{1000}{q^2} + 1000$ The cost function $C(q)=0.2q^2+6q+50$ From the given function for $p$, the revenue function is given by $R(q)=\frac{1000}{q} + 1000q$ The marginal profit is: $P(q)=R(q)-C(q)$ $=\frac{1000}{q} + 1000q-(0.2q^2+6q+50)$ $=-0.2q^2+994q+\frac{1000}{q}-50$ $P'(q)=-0.4q+994-\frac{1000}{q^2}$ When the demand is 10: $P'(10)=-0.4(10)+994-\frac{1000}{(10)^2}=980$