Problem 4 ( 5pts.) - In the past, our revenue function was a function of demand (which in the past was often represented by letter x instead of q.) Now suppose that the demand function is solved for q, that is q=f(p) instead of p=f(q) as usual. We know that revenue function is price times q, the number of items sold. \{Remember again that q is what we called x in the past.\}. So, now we can write R as a function of price, The reason is that revenue is price times quantity sold, that is R(p)=qp or replacing q with f(p), we have R(p)=pf(p). So R is written strictly in terms of price instead of the usual x(=q). Notice that, pf(p) is a product. So if we differentiate it with respect to p we can use the product rule. Let's do it: (' means d/dp, that is differentiate with respect to p ) R(p)=(p)f(p)+f(p)p=1f(p)+pf(p)() Your task, using this formula, is to prove the relationship between elasticity, derivative of revenue, and demand claimed in (a) below. w) Show that R(p)=f(p)[1E] where E is the elasticity of demand. (Hint: f(p) is another notation for dq/dp. In the elasticity formula, solve for f(p) to get f(p)=qE/p. Now substitute this (f(p) in the equation above (marked by ). Factor f(p). Note that q=f(p) so q and f(p) are interchangeable. b) Use the formula you proved in (a) to show that the revenue increases when p increases and demand is inelastic, that is E(p)1 4 Prove that the revenue is maximum when E(p)=1. (Hint: use parts b and c.)