Calculus Question

Question

A company runs food service concessions for sporting events throughout the country. Their marketing research department chose a particular football stadium to test market a new jumbo hot dog. It was found that the demand for the new hot dog is given approximately by

p=8−ln(x),    5≤x≤500,

where x is the number of hot dogs (in thousands) that can be sold during one game at a price of p dollars. If the company pays 1 dollar for each hot dog, how should the hot dogs be priced to maximize the profit per game?

Answer 1

The objective function, Q, is the potential profit that could be made.   From the problem statement

 Q(x)  = p(x) x - 1 x  =   (8 - ln(x)) x -x   = 7x - x ln(x)

 

The calculus approach is to find the value of x that maximizes Q.     This is done by taking the derivative of Q and setting this to zero.

 

   Q'  =   7 - ln(x) -1  =   6 - ln(x)

(To get this result, the product rule has been used)

 

  Setting this equal to zero gives    ln(x) = 6    so  x  =  exp(6) ~ 403.  

Notice that 403  lies between 5 and 500 so is valid.    Plugging exp(6) into p(x)  

gives p = 2,   so the hot dog should be priced at $2.  

 

It should be noted that 403 hot dogs per game might be reasonable, but 403 thousand hot dogs per game is not.

 

 

Answer 2

Hi Kaleb...

 

This is a straight forward max min problem.  By now you should know the process is to take the first derivative and set it equal  to zero.  Then solve that derivative for x. Plug that x value back into original equation.  Check this  P value versus the boudaries set.(5 to 500)

 

Make sure you answer the question by subtracting the dollar paid for the hot dog to get the profit.

 

Good luck