Question

A tour operator charges Rs. 200 per passenger for 50 passengers with a discount of Rs. 5 for every 10 passengers in excess of 50. Determine the number of passengers that will maximize the revenue of the operator.

Answer 1

The number of passengers that will maximize the revenue of the operator is 225 passengers.

Step-by-step explanation:

We are given that a tour operator charges Rs 200 per passenger for 50 passengers with a discount of Rs 5 for every 10 passengers in excess of 50.

AS we know that the revenue of any product is calculated by multiplying the price of that product with the quantity of that product.

So, revenue of operator = Charges per passenger $\times$ Number of passengers

                                         = Rs 200  $\times$  50

                                         = Rs 10,000

But it has been stated by the operator that he will give a discount of Rs 5 for every 10 passengers in excess of 50.

Let the passengers in excess of 50 be '10P' which means that the charges for these passengers will be '200 - 5P'.

Now, the actual revenue of the operator = $(200-5\text{P}) \times (50+10\text{P})$

               R   =  $10000+2000\text{P}-250\text{P}-50 (\text{P})^{2}$

               R   =  $10000+1750\text{P}-50 (\text{P})^{2}$

Now, differentiating the revenue function with respect to P, we get;

              $\frac{dR}{dP}= 0+1750-(50 \times 2) \text{P}$

              $\frac{dR}{dP}= 1750-100 \text{P}$ --------- [Equation 1]

Equating the above expression with 0 to find the value of P;

              $1750-100 \text{P} =0$

                P  =  $\frac{1750}{100}$ = 17.5

Again differentiating the equation 1 with repsect to P;

             $\frac{d^{2} R}{dP^{2} }= 0-100$

                    = -100 < 0

Since $\frac{d^{2} R}{dP^{2} }<0$, this means that at P = 17.5, the revenue function will have its maximum value.

The revenue function =  $(200-5\text{P}) \times (50+10\text{P})$

                                    =  $(200-(5\times 17.5)) \times (50+(10\times17.5))$

                                    =  $112.5 \times 225$ = Rs 25,312.5

Hence, the number of passengers that will maximize the revenue of the operator = $50 + 10\text{P}$  = $50 + (10 \times 17.5)$

               = 50 + 175 = 225 passengers.