A monopolist faces a demand function $Q=4000(p+7)^{-2}$. If she charges a price of p, her marginal revenue will be:
(a) $p/2+ 7$
(b) $2p+ 3.50$
(c) $p/2-7/2$
(d) $-2(p+7)^{-3}$
Correct answer is obviously c). Why is that? Why marginal revenue is in terms of p here? How to calclulate this?
edit: please explain what is wrong with that question?
On
It seems that posers of these problems sometimes delight in expressing the marginal functions in terms of $ \ p \ $ , rather than in terms of $ \ q \ $ which is used in the definitions. On the other hand, it is a reasonable way to do things frequently, since the business has direct control (generally) over the price set, but not over the demand.
One way to approach these problems when we have a demand function $ \ Q(p) \ $ is to use the Chain Rule and the definition for $ \ R(p) \ $ . We can write
$$ \frac{dR}{dp} \ = \ \frac{dR}{dQ} \ \cdot \ \frac{dQ}{dp} \ \ , $$
and with $ \ R(p) \ = \ Q(p) \ \cdot \ p \ $ , we have
$$ \frac{dR}{dp} \ \ = \ \frac{dQ}{dp} \ \cdot \ p \ + \ Q \ \cdot \ 1 \ \ $$
$$ \Rightarrow \ \ \frac{dR}{dQ} \ = \ \frac{\frac{dQ}{dp} \ \cdot \ p \ + \ Q \ \cdot \ 1}{\frac{dQ}{dp}} \ \ = \ \ p \ + \ \frac{Q}{{\frac{dQ}{dp}}} \ \ . $$
For our demand function, we have $ \ {\frac{dQ}{dp}} \ = \ 4000 \ \cdot \ (-2) \ \cdot \ (p+7)^{-3} \ $ , which gives us for the marginal revenue,
$$ \frac{dR}{dQ} \ \ = \ \ p \ + \ \frac{4000 \ \cdot \ (p+7)^{-2}}{{4000 \ \cdot \ (-2) \ \cdot \ (p+7)^{-3}}}$$
$$ = \ \ p \ + \ \frac{1}{-2} \cdot (p+7) \ \ = \ \ p \ - \ \frac{p}{2} \ - \ \frac{7}{2} \ \ = \ \ \frac{p}{2} \ - \ \frac{7}{2} \ \ . $$
It is no coincidence that this looks rather like what we also do to calculate the "price elasticity of demand".
You have $p+7=\sqrt{\frac{4000}{Q}}$. Increasing $Q$ by 1 decreases the price that can be charged (on all sales) by $\Delta p$, where $p-\Delta p+7=\sqrt{\frac{4000}{Q+1}}$. So we have $\Delta p=\sqrt\frac{4000}{Q}-\sqrt{\frac{4000}{Q+1}}=\sqrt{\frac{4000}{Q}}\left(1-\sqrt{\frac{Q}{Q+1}}\right)$ $=(p+7)\left(1-\sqrt{\frac{1}{1+\frac{1}{Q}}}\right)$. We now use the approximation $\frac{1}{\sqrt{1+x}}\approx1-\frac{1}{2}x$ for small $x$. So we get $\Delta p\approx(p+7)\frac{1}{2Q}$.
Hence the loss of revenue from the price fall is $Q\cdot\Delta p=\frac{1}{2}(p+7)$. The extra revenue from the extra sale is $p$. So the net extra revenue is $\frac{1}{2}(p-7)$.