Q. 61 - Calculus

Question

Your family makes and sells velvet Elvis paintings. After many years of research, you have found a function that predicts how many paintings you will sell in a year, based on the price that you charge per painting. You always charge between \(5.00 and \)55.00 per painting. Specifically, if you charge c dollars per painting, then you can sell $N\left(c\right)=0.6c^2-54c+1230$ paintings in a year. 

(a) What price should you charge to sell the greatest number of velvet Elvis paintings, and how many could you sell at that price? For what price would you sell the least number of paintings, and how many would you sell? 

(b) Write down a function that predicts the revenue R(c), in dollars, that you will earn in a year if you charge c dollars per painting. (Hint: Try some examples first; for example, what would your yearly revenue be if you charged \(10.00 per painting? What about \)50.00? Then write down a function that works for all values of c.) 

(c) What price should you charge to earn the most money, and how much money would you earn? What price per painting would cause you to make the least amount of money in a year, and how much money would you make in that case? 

(d) Explain why you do not make the most money at the same price per painting for which you sell the most paintings. 

Step-by-Step Solution

Answer

Part (a) Greatest number of paintings sold is 975 at $\$5$ each and the least number of paintings is 15 at $\$45$ each.

Part (b) $c\left(0.6c^2-54c+1230\right)$

Part (c) Highest earning is $\mathrm{\$}8327.10$

Part (a) Step 1. Given information.

We have been given that per painting costs between $5.00 and $55.00. If the charge per painting is c dollars, then total paintings in a year can be sold as $N\left(c\right)=0.6c^2-54c+1230$.

We have to find the price that should be charged to sell the greatest number of velvet Elvis paintings, and how many can be sold at that price. At what price the least number of paintings can be sold, and how many would be sold.

Part (a) Step 2. Price for painting

Let the cost of the painting is c dollar then the nunmber of painting sold :

$N\left(c\right)=0.6c^2-54c+1230$

Finding the derivative,

$\begin{array}{r}N\left(c\right)=0.6c^2-54c+1230\\ N^{\mathrm{\prime }}\left(c\right)=1.2c-54\\ N^{\ast }\left(c\right)=1.2\end{array}$

Now,

$\begin{array}{r}1.2c-54=0\\ c=45\end{array}$

So the number of painting sold for each values of c are :

$\begin{array}{c}N\left(5\right)=0.6(5{)}^2-54(5)+1230=975\\ N\left(45\right)=0.6(45{)}^2-54(45)+1230=15\\ N\left(55\right)=0.6(55{)}^2-54(55)+1230=\mathrm{\partial }75\end{array}$

Part (b) Step 1. A function that predicts the revenue

The revenue function is:

$R\left(c\right)=c\left(0.6c^2-54c+1230\right)$

Part (c) Step 1. Solve for the price

The revenue function is :

$\begin{array}{r}R\left(c\right)=c\left(0.6c^2-54c+1230\right)\\ =0.6c^3-54c^2+1230c\end{array}$

Finding the derivative,

$\begin{array}{r}R\left(c\right)=0.6c^3-54c^2+1230c\\ R^{\mathrm{\prime }}\left(c\right)=1.8c^2-108c+1230\\ R^{\ast }\left(c\right)=3.6c-108\end{array}$

Now,

$\begin{array}{r}1.8c^2-108c+1230=0\\ c=\frac{90+5\sqrt{78}}{3},\frac{90-5\sqrt{78}}{3}\\ c=44.72,15.28\end{array}$

The total money earned is :

$\begin{array}{r}R\left(44.72\right)=0.6(44.72{)}^3-54(44.72{)}^2+1230\left(44.72\right)=\mathrm{\$}672.90\\ R\left(15.28\right)=0.6(15.28{)}^3-54(15.28{)}^2+1230\left(15.28\right)=\mathrm{\$}8327.10\end{array}$

Part (d) Step 1. Explanation

The highest number of painting were sold by charging $5 per painting but this amount is too low compared to $15.28 because only $5 is earned by selling 1 painting.