I am learning calculus and minima maxima . This is the sum where I I stuck on have no idea

Question

A school charges Rs 50 per student 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

Using a derivative......the maximum revenue occurs when $n > 300$. The equation for revenue is

$R = 50n - 0.1(n - 300)n$...... which simplifies to:

$R = -0.1n^2 + 80n$

Max $R$ occurs where $R' = 0$

$R' = -0.2n + 80 = 0$

$n = \frac{80}{0.2} = 400$

Max $R = -0.1(400)^2 + 80(400) = 16 000$ rupees.

Answer 2

There are 100 "paise" in one "rupee" so 10 paise is 0.1 rupee. If the number of students, x, is larger than 300, the revenue per student is 50- 0.1(x- 300)= 80- 0.1x. The total revenue, then, is $x(80- 0.1x)= 80x- 0.1x^2$. Complete the square: $80x- 0.1x^2= -0.1(x^2- 800x+ 160000- 160000)= -0.1(x- 400)^2+ 16000$. Since that is 16000 minus something, it will be maximum when that "something" is 0- when x= 400.