Let the number of $5 decreases be n
so cost per child = 65 - 5n
number of children = 150 + 10n
Revenue = (65-5n)(150+10n)
= 9750 - 100n - 50n^2
d(Revenue)/dx = -100 -100n = 0 for a max/min of Revenue
100n = -100
n = -1
so the number of decreases is -1, or there should be an increase of
$5 for a daily fee of $70
check:
at $65 , number is 150, R = 9750
at $60, number is 160, R = 9600
at $70, number is 140, R = 9800
at $75, number is 130, R = 9750
how would i do this quadratic word problem
A local daycare centre charges $65 per day to care for a child. The daycare currently cares for 150 children per day. A survey shows that the enrollment in the daycare will increase by 10 children for each $5 decrease in daily fee, and would decrease similarly if the daily fees are increased.
What daily fee would result in the greatest revenue for the daycare centre?
2 answers
x = amount of increase of fee
enrollment = n = 150 - 10 (x/5)
= 150 - 2x
revenue = n * (65+x)
so
revenue = (150-2x)(65+x)
R = 9750 + 20x -2 x^2
find the vertex
R/2 = 4875 +10 x -x^2
x^2 -10 x = -(R/2) + 4875
x^2 - 10 x + 100 = -(R/2) +4975
(x-10)^2 =
so vertex at x = 10
so fee at vertex = 65+10 = 75
enrollment = n = 150 - 10 (x/5)
= 150 - 2x
revenue = n * (65+x)
so
revenue = (150-2x)(65+x)
R = 9750 + 20x -2 x^2
find the vertex
R/2 = 4875 +10 x -x^2
x^2 -10 x = -(R/2) + 4875
x^2 - 10 x + 100 = -(R/2) +4975
(x-10)^2 =
so vertex at x = 10
so fee at vertex = 65+10 = 75
2 answers