Experts please help. Thank you.

Question

Suppose that the number of bacteria in a culture at time t is given by N=2000 (31 + te^t/-50).

(a) Find the largest and smallest number of bacteria in the culture during the time interval 0 < t < 100.

Round your answers to the nearest integer.

Largest number of bacteria: ?

Smallest number of bacteria: ?

(b) At what time during the time interval in part (a) is the number of bacteria decreasing most rapidly?

t = ?

Answer 1

The absolute max or min for the function on [0,100] happens at either the endpoints of the interval or at critical points in the interval, where a critical point means the 1st derivative is equal to zero.

N(t) = 2000(31 + t e^-t/50) = 62000 + 2000te^-t/50

N'(t) = 2000t (e^-t/50)(-1/50) + 2000e^-t/50 = e^-t/50[ -40t + 2000] (product rule)

N'(t) equals zero when -40t + 2000 = 0, or t = 50. Note that e^-t/50 is never equal to zero for any value of t.

Evaluate:

N(0)

N(50)

N(100)

to determine where the absolute max and min values exist on [0, 100].

N(0) = 62000

N(50) = 62000 + 100000/e ≈ 98788

N(100) = 62000 + 200000/e² ≈ 89067

The number of bacteria is decreasing when N'(t) is negative. We know N'(t) = 0,when t = 50.

N'(1) ≈ 1921 > 0 so N(t) is increasing from 0 to 50.

N'(60) ≈ -120 < 0, so N(t) is decreasing from 50 to 100.

The above two calculations show an application of the 1st derivative test.

It is decreasing most rapidly when N'(t) is a minimum. Find N''(t) set it equal to zero.

The derivative of N'(t):

N''(t) = e^-t/50[-80 + (4/5)t ]

N''(t) equals zero, when 4t/5 = 80 or t = 100.

Use the slider on point P to see that the slope of the tangent line to N is most negative (steepest downhill slope) at t = 100:

desmos.com/calculator/pbn8103b76

Slide the graph up a bit to see that N'(t) equals zero when t = 50.