Answer
980
Work Step by Step
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:
$P'(10)=-0.4(10)+994-\frac{1000}{(10)^2}=980$
