College

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ A tray of lasagna comes out of the oven at [tex]$200^{\circ} F$[/tex] and is placed on a table where the surrounding room temperature is [tex]$70^{\circ} F$[/tex]. The temperature [tex]T[/tex] (in [tex]^{\circ} F[/tex]) of the lasagna is given by the function [tex]T(t) = e^{(4.86753-t)} + 70[/tex], where [tex]0 \leq t[/tex] and [tex]t[/tex] is the time (in hours) after taking the lasagna out of the oven.

At which of the following times is the lasagna cooling SLOWEST?

A. 3 hours after being taken out of the oven
B. 5 hours after being taken out of the oven
C. 6 hours after being taken out of the oven
D. 7 hours after being taken out of the oven
E. 9 hours after being taken out of the oven

Answer :

We are given the temperature function for the lasagna as

[tex]$$
T(t) = e^{(4.86753-t)} + 70,\quad t\ge0,
$$[/tex]

where [tex]$t$[/tex] is in hours and [tex]$T(t)$[/tex] is in degrees Fahrenheit. The rate of cooling is determined by the derivative of [tex]$T(t)$[/tex] with respect to [tex]$t$[/tex]. We can compute the derivative as

[tex]$$
T'(t) = \frac{d}{dt} \left(e^{(4.86753-t)} + 70\right).
$$[/tex]

Since the derivative of a constant is zero and using the chain rule for the exponential part, we have

[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]

The quantity [tex]$T'(t)$[/tex] is negative because the temperature is decreasing, and the magnitude of this derivative, [tex]$\left| T'(t) \right| = e^{(4.86753-t)}$[/tex], represents the cooling rate.

To determine when the lasagna cools slowest, we need to find the time (from the provided options) at which [tex]$\left| T'(t) \right|$[/tex] is smallest. This corresponds to the value of [tex]$T'(t)$[/tex] that is closest to zero.

Let’s evaluate the derivative at the given times:

- At [tex]$t=3$[/tex] hours:

[tex]$$
T'(3) = -e^{(4.86753-3)} \approx -6.47229,
$$[/tex]

- At [tex]$t=5$[/tex] hours:

[tex]$$
T'(5) = -e^{(4.86753-5)} \approx -0.87593,
$$[/tex]

- At [tex]$t=6$[/tex] hours:

[tex]$$
T'(6) = -e^{(4.86753-6)} \approx -0.32224,
$$[/tex]

- At [tex]$t=7$[/tex] hours:

[tex]$$
T'(7) = -e^{(4.86753-7)} \approx -0.11854,
$$[/tex]

- At [tex]$t=9$[/tex] hours:

[tex]$$
T'(9) = -e^{(4.86753-9)} \approx -0.01604.
$$[/tex]

Since the absolute value of the derivative represents the rate of cooling, we compare the magnitudes:

[tex]$$
6.47229,\quad 0.87593,\quad 0.32224,\quad 0.11854,\quad 0.01604.
$$[/tex]

The cooling is slowest when the magnitude is the smallest, which occurs when [tex]$t = 9$[/tex] hours.

Thus, the lasagna cools slowest [tex]$9$[/tex] hours after being taken out of the oven.