Differential Calculus "Rate of Change" (Best answer will be voted)

After a bactericide is introduced into a culture of bacteria, the number of bacteria B(t) present at time t is given by the function B(t) = -5t^2 + 50t + 1000, where B(t) is measured in millions and t in hours provided that 0 ≤ t ≤ 20.

(a) Determine the rate of change of the number of bacteria with respect to time at t = 3, t = 5, t = 7. At what time does the bacteria population start to decline?

(b) Find the maximum number of bacteria in the culture.

Show the solution in detailed. Thank you, Best answer will be vote

B(t) = - 5t² + 50t + 1000

a.) Rate of Change: B'(t):

B'(t) = - 10t + 50

t = 3:

B'(3) = - 10(3) + 50
B'(3) = - 30 + 50
B'(3) = 20

At t = 3, rate of change is 20 million/hr.
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t = 5:

B'(5) = -10(5) + 50
B'(5) = - 50 + 50
B'(5) = 0

At t = 5, there is no change in growth rate.
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t = 7:

B'(7) =- 10(7) + 50
B'(7) = - 70 + 50
B'(7) = - 20

At t = 7, there is a decrease in growth of 20 million/hr.
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Bacteria begin to decline at B'(t) = 0:

- 10t + 50 = 0
- 10t = - 50
t = - 50 / - 10
t = 5

Bacteria begin to decline after 5 hrs.
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b.) Maximum bacteria at B(5):

B(5) = - 5(5)² + 50(5) +1000
B(5) = - 5(25) + 250 +1000
B(5) = - 125 + 250 +1000
B(5) = 1125

Maximum No. Bacteria = 1125
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