Question

Mike and Beth are saving money to go to Disneyworld. They need at least $1975 in order to go. Mike mows yards and Beth will wash cars to raise money. Mike charges $25 each time he mows a yard and Beth charges $15 for each car she washes. The number of cars that Beth wash is no more than four times the number of lawns Mike has scheduled to mow. Beth will wash at least 50 cars.Write a set of constraints to model the problem, with x representing the number of lawns mowed and y representing the number of dogs walked.
I'M LOSING TIME PLEASE ANWSER BEFORE ITS TOO LATE

Answer 1

Let x represent the number of lawns mowed and y represent the number of cars washed.

1. Since the number of cars that Beth wash is no more than four times the number of lawns Mike has scheduled to mow, then y≤4x.

2. If Beth will wash at least 50 cars, then y≥50.

3. Mike charges $25 each time he mows a yard, so for x yards he will get $25x. Beth charges $15 for each car she washes, then the total amount of money she earns is $15y. Altogether they get $(25x+15y) that is at least $1975. This means that 25x+15y≥1975.

4. You get three inequalities:

$\left\{\begin{array}{l}y\le 4x\\y\ge 50\\25x+15y\ge 1975\end{array}\right.$