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Zorluk: Çok zorInterpreting Linear Relationships in Context

An industrial oven is used in a bakery. The temperature of the oven chamber, CC, in degrees Fahrenheit (∘F^\circ\text{F}), tt minutes after the heating element is turned on is modeled by the linear equation:

C=18.5t+72C = 18.5t + 72

After a system upgrade, the starting temperature of the oven is 8∘F8^\circ\text{F} warmer, and the rate at which the oven heats up is 20%20\% faster. During a test of the upgraded oven, the heating element is turned on for 1515 minutes, after which the oven is turned off and cools down at a constant rate of 12∘F12^\circ\text{F} per minute. If the cooling process is also linear, which of the following functions models the temperature of the upgraded oven, UU, in degrees Fahrenheit, mm minutes after it is turned off?

  1. A
    U(m)=−12m+357.5U(m) = -12m + 357.5
  2. B
    U(m)=−12m+483.9U(m) = -12m + 483.9
  3. U(m)=−12m+413U(m) = -12m + 413Cevap
  4. D
    U(m)=12m+413U(m) = 12m + 413

Cevap

The function U(m)=−12m+413U(m) = -12m + 413 models the temperature of the upgraded oven, UU, in degrees Fahrenheit, mm minutes after it is turned off.
To find the temperature model during the cooling phase, we must first determine the state of the oven when the cooling begins. The upgraded oven has a starting temperature of 72+8=80∘F72 + 8 = 80^\circ\text{F} and a heating rate of 18.5×1.20=22.2∘F18.5 \times 1.20 = 22.2^\circ\text{F} per minute. After 1515 minutes of heating, the temperature reaches 22.2×15+80=413∘F22.2 \times 15 + 80 = 413^\circ\text{F}. When the oven is turned off at m=0m = 0 minutes, its temperature is 413∘F413^\circ\text{F}, which serves as the y-intercept of the cooling function. Since the temperature decreases at a constant rate of 12∘F12^\circ\text{F} per minute, the rate of change (slope) is −12-12. Therefore, the linear model is the function showing a rate of change of −12-12 and a starting value of 413413.

Adım Adım Çözüm

1
Determine the upgraded starting temperature and heating rate of the oven.
The upgraded starting temperature is 80∘F80^\circ\text{F} and the upgraded heating rate is 22.2∘F22.2^\circ\text{F} per minute.
The original starting temperature of 72∘F72^\circ\text{F} is increased by 8∘F8^\circ\text{F} to get 72+8=80∘F72 + 8 = 80^\circ\text{F}. The original heating rate (slope) of 18.5∘F18.5^\circ\text{F} per minute is increased by 20%20\%, which is calculated as 18.5×1.20=22.2∘F18.5 \times 1.20 = 22.2^\circ\text{F} per minute.
2
Calculate the temperature of the upgraded oven at the moment it is turned off.
The temperature is 413∘F413^\circ\text{F} at t=15t = 15 minutes.
The heating phase is modeled by the linear relationship H(t)=22.2t+80H(t) = 22.2t + 80. Substituting t=15t = 15 yields H(15)=22.2(15)+80=333+80=413∘FH(15) = 22.2(15) + 80 = 333 + 80 = 413^\circ\text{F}.
3
Construct the linear function for the cooling phase.
U(m)=−12m+413U(m) = -12m + 413
When the oven is turned off (m=0m = 0), its temperature is 413∘F413^\circ\text{F}. Since it cools down at a constant rate of 12∘F12^\circ\text{F} per minute, the slope of the cooling function is −12-12. Thus, the linear model is U(m)=−12m+413U(m) = -12m + 413.

Anahtar Kavram

Interpreting and modifying parameters of linear models in multi-stage contextual scenarios.

Alternatif Yöntem

Instead of writing the heating function explicitly, you can calculate the total temperature increase directly: the temperature rises by 18.5×1.20=22.2∘F18.5 \times 1.20 = 22.2^\circ\text{F} per minute for 1515 minutes, which is a total increase of 22.2×15=333∘F22.2 \times 15 = 333^\circ\text{F}. Adding this increase to the upgraded starting temperature of 72+8=80∘F72 + 8 = 80^\circ\text{F} gives the peak temperature of 80+333=413∘F80 + 333 = 413^\circ\text{F}. Since the cooling phase is linear with a slope of −12-12 and a y-intercept of 413413, the function is immediately determined.
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Interpreting Linear Relationships in Context

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