Question

Many businesses know that most consumers associate price with quality (i.e. if it costs more it mustbe of a higher quality). An actuary at a cosmetics manufacture determined that the profit made theprice charged for lipstick at a particular retail outlet could be modeled by:V(p) =-0.5^3 + 12p^2 - 17p, where p is the price in dollarsV(p) =- 17p, where p is the price in dollars and V(p) is the amount earned by theretail store selling just the lipstick in a month.How much should the store charge to maximize its profit?

Answer 1

Given:

$V(p)=-0.5p^3+12p^2-17p$

Take the derivative with respect to p,

$\begin{gathered} \frac{dV}{dp}=\frac{d}{dp}(-0.5p^3+12p^2-17p) \\ =-\frac{d}{dp}(0.5p^3)+\frac{d}{dp}(12p^2)-\frac{d}{dp}(17p) \\ =-1.5p^2+24p-17 \end{gathered}$

Now set the derivative equal to zero,

$\begin{gathered} \frac{dV}{dp}=0 \\ -1.5p^2+24p-17=0 \\ p=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=-1.5,b=24,c=-17 \\ p=\frac{-24\pm\sqrt[]{24^2-4(-1.5)(-17)}}{2(-1.5)} \\ p=\frac{-24\pm\sqrt[]{474}}{-3} \\ p=\frac{-24+\sqrt[]{474}}{-3},p=\frac{-24-\sqrt[]{474}}{-3} \\ p=\frac{24-\sqrt[]{474}}{3},p=\frac{24+\sqrt[]{474}}{3} \\ p=0.7428,p=15.2572 \end{gathered}$

Now, take second derivative of the function,

$\begin{gathered} V^{\doubleprime}(p)=\frac{d}{dp}(-1.5p^2+24p-17) \\ =-3p+24 \\ V^{\doubleprime}(0.7428)=-3(0.7428)+24=21.7716>0 \\ V^{\doubleprime}(15.2572)=-3(15.2572)+24=-21.7716<0 \end{gathered}$

As the second derivative is less than 0 when p=15.2572 that is maximum profit.

So, the price charge to maximize its profit is p=15.2572 ( approximated)