The problem given is as follows:4. The demand for face masks is given by the equation $ p = -3 \ln(x) + 10 $, with the domain $[1, 28]$. In this equation, $ x $ represents the number of face masks (in thousands) that can be sold at a price of $ p $ dollars. The task is to determine the pricing of the face masks in order to maximize the revenue.This is an application of maximizing a function within the context of economics. To solve the problem, you would typically set up the revenue function $ R(x) = x \cdot p(x) $, where $ p(x) = -3 \ln(x) + 10 $, and find its maximum over the given domain. No graphs or diagrams are present in the image.

Given: The demand for face masks is given by, p=-3 lnx+10, with domain 1, 28 where x is the number of face masks in thousands that can be sold at a price of $p. To Find: How should the face masks be priced in order to maximize the revenue? Solution: The demand for face masks is given by, p=-3 lnx+10, with domain 1, 28 Where, x is the number of face masks in thousands that can be sold at a price of $p. Since, Revenue R=Quantity sold×price⇒R=xp⇒R=x-3 lnx+10⇒R=-3x lnx+10xDifferentiating with respect to 'x' we getdRdx=ddx-3x lnx+10x⇒dRdx=ddx-3x lnx+ddx10x⇒dRdx=-3ddxx lnx+10ddxx Using, ddxc.fx=cddxfx⇒dRdx=-3xddxlnx+lnxddxx+10ddxx Using, ddxu.v=uddxv+vddxu⇒dRdx=-3x1x+lnx1+101 Since, ddxlnx=1x, and, ddxx=1⇒dRdx=-31+lnx+10⇒dRdx=-3-3 lnx+10⇒dRdx=7-3 lnxAgain differentiating with respect to 'x' we getddxdRdx=ddx7-3 lnx⇒d2Rdx2=ddx7-ddx3 lnx⇒d2Rdx2=ddx7-3ddxlnx⇒d2Rdx2=0-31x ⇒d2Rdx2=-3xFor stationary points we have,dRdx=0⇒7-3 lnx=0⇒lnx=73⇒x=e73 Since, x∈1, 28 Now, at x=e73, we haved2Rdx2x=e73=-3e73<0 Since, e73 is always positive.Since, d2Rdx2x=e73<0Hence, The revenue is maximum at x=e73Now, substituting, x=e73, in p=-3 lnx+10, we getp=-3 lne73+10⇒p=-373 ln e+10⇒p=-3×73ln e+10⇒p=-7×1+10 Since, lne=1⇒p=-7+10⇒p=$3⇒The price of face masks=$3Hence, The face masks should be priced of $3 in order to maximize the revenue.