Science:Math Exam Resources/Courses/MATH104/December 2014/Question 03 (b)/Solution 1

Substituting $q=3$ in the relation given, we obtain that $p^{3}=38-3-3^{3}=35-27=8\Rightarrow p=2$. Next, we determine the value of $q'$ when $p=2,q=3$. From the calculation in the above, we know that $q'={\frac {-3p^{2}}{1+3q^{2}}}$, substituting $p=2,q=3$ yields: $q'={\frac {-3}{7}}$. Last, we check what the change in revenue is when $p=2,q=3$. Recall that $R=pq$ and hence, $R'=q+pq'$ using the product rule when differentiating with respect to price, $p$. Substituting $p=2,q=3,q'={\frac {-3}{7}}$, we obtain that $R'=3+2({\frac {-3}{7}})=3-{\frac {6}{7}}>0$. Since $R'>0$ when price $p=2$, slightly raising the price will increase the revenue.