A tray of lasagna comes out of the oven at 200°F and is placed on a table where the surrounding room temperature is 70°F. The temperature T (in °F) of the lasagna is given by the function T(t)=e^94.86753-t)+70 , 0 ≤ t, where t is time (in hours) after taking the lasagna out of the oven.
At which of the following times is the lasagna cooling slowest?
After 9 hours
To determine at which time the lasagna is cooling slowest, we need to find the point on the graph where the slope is the smallest or the derivative is closest to zero.
The temperature function is given by T(t) = e^(94.86753-t) + 70.
Taking the derivative of T(t) with respect to t, we get:
dT(t)/dt = -e^(94.86753-t)
To find the point where the derivative is closest to zero, we set dT(t)/dt = 0:
-e^(94.86753-t) = 0
Since e raised to any power is never zero, there is no time at which the derivative is exactly zero. This means the lasagna is cooling at a constant rate throughout the entire cooling process. Therefore, there is no specific time at which the lasagna is cooling slowest.
To determine at which time the lasagna is cooling slowest, we need to find the minimum value of the derivative of the temperature function T(t).
The derivative of T(t) with respect to t can be found by taking the derivative of each term separately:
T'(t) = -e^(-94.86753-t)
To find the minimum value of T'(t), we need to set T'(t) equal to zero and solve for t:
0 = -e^(-94.86753-t)
To solve this equation, we can take the natural logarithm (ln) of both sides:
ln(0) = ln(-e^(-94.86753-t))
Since the natural logarithm of zero is undefined, there are no real solutions for this equation. Therefore, there is no time at which the lasagna is cooling slowest based on the given temperature function.