Question

The production constraints for a toy car and truck company are given by the equations and graph of the system of inequalities. Production Constraints:Wheels: The company has no more than 360 wheels in stock. Each toy car needs 4 wheels, and each toy truck needs 6 wheels. Seats: The company has no more than 100 seats in stock. Each toy car needs 2 seats, and each toy truck needs 1 seat. 4x + 6y ≤ 360 2x + y ≤ 100 Which does NOT represents a viable option for the companies production of toy cars and trucks?

A) (10, 50) → 10 toy cars and 50 toy trucks
B) (20, 40) → 20 toy cars and 40 toy trucks
C) (30, 40) → 30 toy cars and 40 toy trucks
D) (40, 30) → 40 toy cars and 30 toy trucks

Answer 1

Upon evaluating the production constraints, Option C (30 toy cars and 40 toy trucks) is identified as the non-viable option because it requires more wheels than the company has in stock.

Step-by-step explanation:

The question presented asks to identify a non-viable production option for a toy car and truck company based on the given constraints of wheels and seats availability. We have two constraints for the production:

• Wheels: 4x + 6y ≤ 360
• Seats: 2x + y ≤ 100

By plugging the options A) (10, 50), B) (20, 40), C) (30, 40), and D) (40, 30) into the system of inequalities, we can check their viability:

• Option A: 4(10) + 6(50) ≤ 360 and 2(10) + (50) ≤ 100
• Option B: 4(20) + 6(40) ≤ 360 and 2(20) + (40) ≤ 100
• Option C: 4(30) + 6(40) ≤ 360 and 2(30) + (40) ≤ 100 (This option exceeds the number of available wheels)
• Option D: 4(40) + 6(30) ≤ 360 and 2(40) + (30) ≤ 100

Only Option C results in the use of more wheels than available, making it the non-viable option. Therefore, the option that does NOT represent a viable choice for the company's production plan due to the constraints on wheel usage is C) 30 toy cars and 40 toy trucks.