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Zorluk: ZorInterpreting Linear Relationships in Context

A commercial cargo aircraft's total weight WW, in kilograms, is modeled as a linear function of the volume of fuel FF, in liters, in its fuel tanks. The equation modeling this relationship is:

W=0.8F+74,000W = 0.8F + 74,000

To comply with runway safety regulations at a destination airport, the aircraft's total weight must not exceed 85,00085,000 kilograms upon landing. The aircraft takes off with 18,00018,000 liters of fuel and consumes fuel at a constant rate of 1,7001,700 liters per hour of flight. What is the minimum number of hours the aircraft must fly before it can safely land?

Cevap: 2.5 hours

Cevap

The minimum number of hours the aircraft must fly is 2.5.
To satisfy runway safety regulations, the aircraft's weight must be at most 85,00085,000 kg. According to the weight model W=0.8F+74,000W = 0.8F + 74,000, we set 0.8F+74,000≤85,0000.8F + 74,000 \leq 85,000, which simplifies to 0.8F≤11,0000.8F \leq 11,000. Dividing by the density coefficient 0.80.8 yields F≤13,750F \leq 13,750 liters as the maximum amount of fuel the aircraft can contain at landing. Since the aircraft takes off with 18,00018,000 liters of fuel and consumes 1,7001,700 liters per hour, the fuel remaining after tt hours is 18,000−1,700t18,000 - 1,700t. To ensure the remaining fuel is less than or equal to 13,75013,750 liters, we solve 18,000−1,700t≤13,75018,000 - 1,700t \leq 13,750, which simplifies to 4,250≤1,700t4,250 \leq 1,700t, or t≥2.5t \geq 2.5 hours. Therefore, the minimum duration of the flight is 2.5 hours.

Adım Adım Çözüm

1
Formulate the weight limit inequality using the linear relationship.
0.8F+74,000≤85,0000.8F + 74,000 \leq 85,000
The aircraft's weight WW is given by 0.8F+74,0000.8F + 74,000, where 0.80.8 represents the density of the fuel in kg/L and 74,00074,000 represents the weight of the empty aircraft and its cargo. This total weight must not exceed 85,00085,000 kg.
2
Calculate the maximum fuel capacity allowed at landing by solving the inequality.
F≤13,750F \leq 13,750
Subtracting 74,00074,000 from both sides yields 0.8F≤11,0000.8F \leq 11,000. Dividing by 0.80.8 gives F≤13,750F \leq 13,750 liters as the fuel ceiling for landing.
3
Use the fuel consumption rate to find the minimum flight time.
t≥2.5t \geq 2.5
With an initial fuel volume of 18,00018,000 liters and a burn rate of 1,7001,700 liters/hour, the fuel remaining after tt hours is 18,000−1,700t18,000 - 1,700t. Setting this expression to be at most 13,75013,750 liters yields 18,000−1,700t≤13,75018,000 - 1,700t \leq 13,750. Subtracting 18,00018,000 gives −1,700t≤−4,250-1,700t \leq -4,250, and dividing by −1,700-1,700 results in t≥2.5t \geq 2.5 hours.

Anahtar Kavram

Interpreting linear coefficients in context and setting up linear inequalities with rates to model real-world constraints.
Tahmini Süre:2m 30s
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Interpreting Linear Relationships in Context

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