Suppose the temperature in degrees Celsius over a 10​-hour period is given by T(t)=-t^2+3t+40 . ​a) Find the average temperature. ​b) Find the minimum temperature. ​c) Find the maximum temperature.

Answer 1

take the derivative, and set equal to zero

-2t +3 = 0

t = 3/2 = 1.5 hours

-t^2 +3t + 40 = -(1.5)^2 +3(1.5) + 40 = = 42.25 degrees Celsius= maximum temperature =108.05 degrees Fahrenheit

-459.67 degrees Fahrenheit = absolute zero = -273.15 C = 0 K, an absolute lowest temperature

in 10 hours T = -100+30+40 = -30 degrees Celsius = minimum temperature in a 10 hour period

-273.15 = -t^2 + 3t + 40

t^2 -3t - 313.15 = 0

t =6/2 + (1/2)sqr(9+4(313.15) = about 19.6 hours, when temperature reaches absolute zero = -273.15 C or 0 K

After 19 hours & 36 minutes the temperatures never changes, it stays at absolute zero, where nothing moves and it can never get colder.

t T

0 40

1 42

1.5 42.25

2 42

3 40

4 36

5 30

5.5 26.25

6 22

7 12

8 0

9 -14

10 -30

19 -264

19.6 -273.15

100 -273.15

1000 -273.15

infinity -273.15

average temperature from 0 to infinity is -273.15 C

T(t) = -t^2 +3t + 40 is a downward opening parabola with vertex = (1.5, 42 1/4) = maximum point

minimum point is (19.6, -273.15)

median temperature in a 10 hour period = 26 1/4 C