SOLUTION: 1. Imagine that you have an elementary school-age child and are researching childcare options. You find two that you like - one at a child care center, one is based in a familys ho

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: 1. Imagine that you have an elementary school-age child and are researching childcare options. You find two that you like - one at a child care center, one is based in a familys ho      Log On


   



Question 1116972: 1. Imagine that you have an elementary school-age child and are researching childcare options. You find two that you like - one at a child care center, one is based in a familys home - so it basically comes down to which one will be cheapest. the childcare charges $300 to enroll, then $79 per week. the family-run childcare charges $180 to enroll, then $93 per week.
a) Setup two functions to represent the cost of each.
b)which childcare approach will be the cheapest in the short-term?
c)determine the number of weeks at which it becomes more cost-effective to choose the long-term ?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
c = cost for child care.
h = cost for home

your equations are:

c = 300 + 75w

h = 180 + 93w

w is the number of weeks.

your break even point is when c = h.

that happens when 300 + 79w = 180 + 93w

solve for w to get 14w + 120

solve for w to get w = 120/14 = 8.571428571

if you choose 8 weeks, the cost for each would be:

c = 300 + 79 * 8 = 932
h = 180 + 93 * 8 = 924

therefore, h would be cheaper.

if you choose 9 weeks, the cost for each would be:

c = 300 + 79 * 9 = 1011
h = 180 + 93 * 9 = 1017

therefore, c would be cheaper.

you could also look at it the following way.

break even is when w = 8.571428571

since the incremental cost per week for c is less than the incremental cost per week for h, then c will be cheaper if the number of weeks are greater than the break even number of weeks, and h will be cheaper if the number of weeks are less than the break even number of weeks.

you could also have use an inequality instead of an equation.

you want to know when c is less than h.

since c = 300 + 79 * w and h = 180 + 93 * w,then your inequality becomes:

300 + 79 * w < 180 + 93 * w

subtract 79 * w from both sides of this inequality and subtract 180 from both sides of this inequality to get 300 - 180 < 93 * w - 79 * w

simplify to get 120 < 14 * w

divide both sides of this inequality by 14 to get 120/14 < w

simplify to get 8.571428571 < w

this is the same as w > 8.571428571.

c will be less than h when w > 8.571428571.

consequently c will be greater than h when w < 8.571428571.

you could solve for c > h directly by using the following inequality.

c > h implies 300 + 79 * w > 180 + 93 * w

subtract 180 from both sides of this inequality and subtract 79 from both sides of this inequality to get 300 - 180 > 93 * w - 79 * w.

simplify to get 120 > 14 * w

divide both sides of this inequality by 14 to get 120/14 > w

simplify to get 8.571428571 > w

this is the same as w < 8.571428571

c will be greater than h when w < 8.571428571.