Question

this question has two parts. first, answer part a. then, answer part b.
part a
business for maximum efficiency, a factory must have at
least 100 workers, but no more than 200 workers on a
shift. the factory also must manufacture at least 30 units
per worker.
a. let x be the number of workers and let y be the number
of units. write a system of inequalities expressing the
conditions in the problem.
$x \leq \square$
$x \geq \square$
$y \geq \square x$
part b
b. graph the systems of inequalities on a separate sheet of
paper.
c. select three possible solutions.
□ a) (150, 5100)
□ b) (150, 4000)
□ c) (110, 3410)
□ d) (200, 3000)
□ e) (180, 6300)

Explanation:

Step1: Define worker upper limit

The factory has max 200 workers.
$x \leq 200$

Step2: Define worker lower limit

The factory has min 100 workers.
$x \geq 100$

Step3: Define unit production limit

Each worker makes at least 30 units.
$y \geq 30x$

Step4: Verify option A

Check if $(150, 5100)$ fits all inequalities:
$100 \leq 150 \leq 200$; $5100 \geq 30(150)=4500$: Valid

Step5: Verify option B

Check if $(150, 4000)$ fits all inequalities:
$100 \leq 150 \leq 200$; $4000 < 30(150)=4500$: Invalid

Step6: Verify option C

Check if $(110, 3410)$ fits all inequalities:
$100 \leq 110 \leq 200$; $3410 \geq 30(110)=3300$: Valid

Step7: Verify option D

Check if $(200, 3000)$ fits all inequalities:
$100 \leq 200 \leq 200$; $3000 < 30(200)=6000$: Invalid

Step8: Verify option E

Check if $(180, 6300)$ fits all inequalities:
$100 \leq 180 \leq 200$; $6300 \geq 30(180)=5400$: Valid

Answer:

Part A

$x \leq 200$
$x \geq 100$
$y \geq 30x$

Part B

c. Possible Solutions:

A) (150, 5100)
C) (110, 3410)
E) (180, 6300)

Note: For Part B b., to graph the system:

1. Draw vertical lines $x=100$ (solid, shade right) and $x=200$ (solid, shade left).
2. Draw the line $y=30x$ (solid, shade above the line).
3. The feasible region is the overlapping shaded area between $100 \leq x \leq 200$ and above $y=30x$.