A lawn care store makes a profit of $89.00 for every push mower sold and $236.00 for every riding lawn mower sold. The owner would like to make at least $621.00 a day from the sales of push and riding lawn mowers. Let x = the number of push mowers sold. Let y = the number of riding mowers sold. Which of the following is a list of 3 possible solutions to the graph?(1 point)

Question

A lawn care store makes a profit of $89.00 for every push mower sold and $236.00 for every riding lawn mower sold.  The owner would like to make at least $621.00 a day from the sales of push and riding lawn mowers.  Let x = the number of push mowers sold. Let y = the number of riding mowers sold.  Which of the following is a list of 3 possible solutions to the graph?(1 point)
Responses

(3,1), (4,2), (1,1)
(3,1), (4,2), (1,1)

(2,1), (2,2), (2,3)
(2,1), (2,2), (2,3)

(3,2), (2,3), (1,4)
(3,2), (2,3), (1,4)

(1,2), (2,2), (3,1)

GPT 3.5 · Answer

To solve this problem, we need to find combinations of x and y that satisfy the equation 89x + 236y ≥ 621.

Option 1: (3,1), (4,2), (1,1)
89(3) + 236(1) = 297 + 236 = 533 < 621
89(4) + 236(2) = 356 + 472 = 828 > 621
89(1) + 236(1) = 89 + 236 = 325 < 621

Option 2: (2,1), (2,2), (2,3)
89(2) + 236(1) = 178 + 236 = 414 > 621
89(2) + 236(2) = 178 + 472 = 650 > 621
89(2) + 236(3) = 178 + 708 = 886 > 621

Option 3: (3,2), (2,3), (1,4)
89(3) + 236(2) = 267 + 472 = 739 > 621
89(2) + 236(3) = 178 + 708 = 886 > 621
89(1) + 236(4) = 89 + 944 = 1033 > 621

Option 4: (1,2), (2,2), (3,1)
89(1) + 236(2) = 89 + 472 = 561 < 621
89(2) + 236(2) = 178 + 472 = 650 > 621
89(3) + 236(1) = 267 + 236 = 503 < 621

Out of the given options, option 2: (2,1), (2,2), (2,3) is the only one where all combinations satisfy the equation. Therefore, the correct answer is option 2: (2,1), (2,2), (2,3).