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

From the previous part, we know that when $q=3$, by the relation give, $p=2$. Now, we take that above relation and differentiate it with respect to time: $3p^{2}{\frac {dp}{dt}}+{\frac {dq}{dt}}+3q^{2}{\frac {dq}{dt}}=0$ We know that ${\frac {dp}{dt}}=7,p=2,q=3$. Substitute into the new found relation and isolate for ${\frac {dq}{dt}}$ gives:

${\frac {dq}{dt}}={\frac {-3(2)^{2}(7)}{1+3(3)^{2}}}={\frac {-84}{28}}=-3$.

So the demand is decreasing at a rate of $3$ units/month.