Question

A school charges Rs 50 per student for transport facilities if not more than 300 students enroll.The charges decrease by 10 paise for each student in excess of 300.How many students should be enrolled so as to have maximum revenue?

Answer 1

400

Answer 2

Explanation of the solution:

1. Define Variables:
   Let $N$ be the total number of students enrolled.
   Let $C$ be the charge per student (in Rupees).
   Let $R$ be the total revenue.

2. Formulate Charge Function:

   • If $N \le 300$, the charge is fixed at Rs 50.
   • If $N > 300$, the charge decreases by 10 paise (Rs 0.10) for each student in excess of 300.
   • Let $x$ be the number of students in excess of 300. So, $x = N - 300$.
   • The charge per student $C$ for $N > 300$ is given by:
     $C = 50 - 0.10x$
     Substitute $x = N - 300$:
     $C = 50 - 0.10(N - 300)$
     $C = 50 - 0.10N + 30$
     $C = 80 - 0.10N$

3. Formulate Revenue Function:
   The total revenue $R(N)$ is the product of the total number of students and the charge per student:
   $R(N) = N \times C$
   For $N > 300$:
   $R(N) = N(80 - 0.10N)$
   $R(N) = 80N - 0.10N^2$

4. Find Maximum Revenue using Calculus:
   To find the number of students that maximizes revenue, we differentiate $R(N)$ with respect to $N$ and set the derivative to zero.
   $\frac{dR}{dN} = \frac{d}{dN}(80N - 0.10N^2)$
   $\frac{dR}{dN} = 80 - 0.20N$

   Set $\frac{dR}{dN} = 0$:
   $80 - 0.20N = 0$
   $0.20N = 80$
   $N = \frac{80}{0.20}$
   $N = \frac{80}{\frac{2}{10}}$
   $N = 80 \times \frac{10}{2}$
   $N = 40 \times 10$
   $N = 400$

5. Verify Maximum:
   To confirm this is a maximum, we can use the second derivative test:
   $\frac{d^2R}{dN^2} = \frac{d}{dN}(80 - 0.20N)$
   $\frac{d^2R}{dN^2} = -0.20$
   Since $\frac{d^2R}{dN^2} = -0.20 < 0$, the value $N = 400$ corresponds to a local maximum.

6. Check Domain Validity:
   The formula for $R(N)$ was derived for $N > 300$. Our calculated value $N = 400$ satisfies this condition.
   If $N \le 300$, the revenue is $R(N) = 50N$. The maximum in this range occurs at $N=300$, giving $R(300) = 50 \times 300 = 15000$.
   For $N=400$, the revenue is $R(400) = 80(400) - 0.10(400)^2 = 32000 - 0.10(160000) = 32000 - 16000 = 16000$.
   Since $16000 > 15000$, the maximum revenue is indeed achieved at $N=400$.

The number of students must be an integer, and since our result is exactly 400, no further adjustment is needed.