SOLUTION: I have having a problem trying to figure out a new equation for my problem. Here is the set up. A father is looking for the best option for daycare for his 5 year old son. A ho

Question 855265: I have having a problem trying to figure out a new equation for my problem. Here is the set up.
A father is looking for the best option for daycare for his 5 year old son. A home-based option (Option A) charges a flat rate per hour of $15. A center-based option (option B) charges a fixed fee per week of $150 for 40 hours and $10 an hour for any hour over the 40. What is the best option for the father?
The father works 50 hours a week from his office which is located in his home. Both childcare options are equal distance from his house so the driving expenses don�t need to be considered. Both child care options offer a two week vacation time without having to pay child care but still hold the child�s spot in the daycare. This works well with the father who also gets two weeks of vacation time a year.
Here are my two equations.
Option A: (A)y = 15x
Option B1: (B1)y= 150 + 10 (x-40) � day care with 50 hour child care week
But my paper got sent back saying I need to come up with an additional equation for Option B that represents less than 40 hours of daycare is needed to complete the assessment. I am not sure how to set this up. I came up with this.
y=150+10(x-40), x≥40 hours
Is that right??? If so how do prove when the two equations will be equal.
15x=150+10(x-40), x≥40 hours????
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
There is obviously something wrong with the charge of $15 an hour.
We are told the childcare will be needed for 50 hours a week. There is no reason to assess less than 40 hours.
50 hours
y = 15*50=750
y= 150 + 10 (10)=250
less than 40
y=15*30=450

y=150
The plans are equal for ten hours. After ten hours Plan B is better
Plan A is better for less than ten hours but is not an option for the father.

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