A computer store is placing an order for computers and tablets. Each computer costs $875, and each tablet costs $235. The store can spend at most $10,000 for these computers and tablets. The computer store has a limited amount of space for displaying the new computers and tablets. Each computer takes up 4 square feet of space, and each tablet takes up 3 square feet of space. The store has only 45 square feet of display space available for the new computers and tablets. The system of inequalities shown can be used to determine possible combination of computers, x, and tablets, y, the store can order.
875x + 235y ≤ 10,000
4x +3y ≤ 45
Which combinations of computers, x, and tablets, y, can the computer store order?
(7,9)
(10,2)
(8,4)
(5,9)
To determine which combinations of computers and tablets the store can order, we need to solve the system of inequalities.
The first inequality is 875x + 235y ≤ 10,000, which represents the budget constraint. It states that the cost of the computers (875x) plus the cost of the tablets (235y) must be less than or equal to $10,000.
The second inequality is 4x + 3y ≤ 45, which represents the space constraint. It states that the total square footage used by the computers (4x) plus the total square footage used by the tablets (3y) must be less than or equal to 45 square feet.
Let's evaluate each combination of computers (x) and tablets (y) using these inequalities:
(7, 9):
875(7) + 235(9) = 6,125 + 2,115 = 8,240 (budget constraint)
4(7) + 3(9) = 28 + 27 = 55 (space constraint)
Since 8,240 > 10,000, this combination does not meet the budget constraint.
(10, 2):
875(10) + 235(2) = 8,750 + 470 = 9,220 (budget constraint)
4(10) + 3(2) = 40 + 6 = 46 (space constraint)
Since 9,220 < 10,000 and 46 > 45, this combination meets both the budget and space constraints.
(8, 4):
875(8) + 235(4) = 7,000 + 940 = 7,940 (budget constraint)
4(8) + 3(4) = 32 + 12 = 44 (space constraint)
Since 7,940 < 10,000 and 44 < 45, this combination meets both the budget and space constraints.
(5, 9):
875(5) + 235(9) = 4,375 + 2,115 = 6,490 (budget constraint)
4(5) + 3(9) = 20 + 27 = 47 (space constraint)
Since 6,490 < 10,000 and 47 > 45, this combination meets both the budget and space constraints.
Therefore, the computer store can order the combinations (10, 2), (8, 4), and (5, 9).