The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
Explanation
Let P represent the cost ; n, the number of pupils and k and c, constant terms.
(therefore P = k + cn)
(15705 = k + 50c … (1))
(13305 = k + 40c …. (2))
(1) – (2) :
(2400 = 10c implies c = frac{2400}{10} = 240)
Put c = 240 in (1), so that
(15705 = k + (50)(240))
(15705 = k + 12000 implies k = 3705)
(P = 3705 + 240n) —— formula connecting price and number of pupils.
(a) When number of pupils, n = 44
(P = 3705 + 240(44) = 3705 + 10560 = $14265)
(b) For school not to be at loss:
(360 geq k + cn)
(360n geq 3705 + 240n)
(360n – 240n geq 3705 implies 120n geq 3705)
(n geq frac{3705}{120})
(n geq 30.875 )
(n geq 31)
The pupils should be 31 at least for the school not to run at loss.