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Question

Elisa Dimas is the manager for the Learning Loft Day Care Center. The center offers all day service for pre-school children for $18 per day and after school only service for $6 per day. Fire codes permit only 50 children in the building at one time. State law dictates that a child care worker can be responsible for a maximum of 3 pre-school children and 5 school age children at one time. Ms. Dimas has ten child care workers available to work at the center during the week. How many children of each age group should Ms. Dimas accept to maximize the daily income of the center?

Answer 1

Hi,

Let x = number of all-day pre-school children
Let y = number of after school children
x/3 = number of child care workers needed for pre-school children
y/5 = number of child care workers needed for after school children

x + y ≤ 50
x/3 + y/5 ≤ 10

Multiply the second inequality by 15 to eliminate the fractions.
15(x/3 + y/5 ≤ 10)
5x + 3y ≤ 150

Solve both inequalities for y. Graph those expressions and find the points of intersections of the shaded area.
y ≤ -x + 50
y ≤ -5/3x + 50
y ≥ 0 (There can't be less than 0 students.)
x ≥ 0 (There can't be less than 0 students.)

The shaded area is a triangle with vertices at (0,50), (30,0) and (0,0). To find the maximum income of the center, plug these values into the cost formula, 18x + 6y.
(0,50) provides 18(0) + 6(50) = $300 income
(30,0) provides 18(30) + 6(0) = $540 income
(0,0) provides $0 income.

Having 30 pre-school children all day and no after-school children will provide the maximum income for the center.

I hope that helps!! :-)