Wednesday

April 16, 2014

April 16, 2014

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

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 10:32pmLet 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

- math -
**Damon**, Wednesday, March 17, 2010 at 10:33pmx = 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

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