Posted by jake on Wednesday, March 17, 2010 at 9:44pm.
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?

math  Reiny, Wednesday, March 17, 2010 at 10:32pm
Let the number of $5 decreases be n
so cost per child = 65  5n
number of children = 150 + 10n
Revenue = (655n)(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

math  Damon, Wednesday, March 17, 2010 at 10:33pm
x = amount of increase of fee
enrollment = n = 150  10 (x/5)
= 150  2x
revenue = n * (65+x)
so
revenue = (1502x)(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
(x10)^2 =
so vertex at x = 10
so fee at vertex = 65+10 = 75
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