Question

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. 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 combinations of computers, x, and tablets, y, the store can order. 875x + 235y ≤ 10,000 4x + 3y ≤ 45 which combination of computers, x, and tablets, y, can the store order? a (8,4) b (5,9) c (10,2) d (7,9)

Explanation:

Step1: Check space - constraint for option A

For the space - constraint $4x + 3y\leq45$, when $x = 8$ and $y = 4$, we have $4\times8+3\times4=32 + 12=44\leq45$. For the cost - constraint $875x + 235y\leq10000$, we have $875\times8+235\times4=7000+940 = 7940\leq10000$.

Step2: Check space - constraint for option B

When $x = 5$ and $y = 9$, $4x+3y=4\times5 + 3\times9=20+27 = 47>45$. So option B does not satisfy the space - constraint.

Step3: Check space - constraint for option C

When $x = 10$ and $y = 2$, $4x+3y=4\times10+3\times2=40 + 6=46>45$. So option C does not satisfy the space - constraint.

Step4: Check space - constraint for option D

When $x = 7$ and $y = 9$, $4x+3y=4\times7+3\times9=28+27 = 55>45$. So option D does not satisfy the space - constraint.

Answer:

A. (8,4)