Question

The number of fish in a small bay is modeled by the function F defined by F(t)=10 (t3 â 12t2 + 45t +100), where t is measured in days and 0 (a) Using correct units, interpret the meaning of F' (4)= -30 in the context of the problem.
(b) Based on the model, what is the absolute minimum number of fish in the bay over the time interval 0 (C) For what values of t, 0 (d) For 0

Answer 1

The graph of the given function is attached showing characteristics of the function.

The correct responses are;

• (a) F'(4) = -30 means that on day 4, the number of fish in the bay is decreasing at a rate of 30 fish per day.

• (b) Over the interval 0 ≤ t ≤ 8 the absolute minimum number of fish in the bay is; 1,000.

• (c) The values of t at which the rate of change is decreasing is; 3 ≤ t ≤ 8

• (d) The rate of change of the number of pelicans flying near is; P' =  10·(3·c² - 24·c  + 45)

Reasons:

The given function is; f(t) = 10·(t³ - 12·t² + 45·t + 100)

Where;

t = Number of days

0 ≤ t ≤ 8

(a) The derivative of the given function is presented as follows;

F'(t) = 10 × (3·t² - 2×12·t + 45) = 10·(3·t² - 24·t  + 45)

Therefore;

F'(4) = 10 × (3 × 4² - 24 × 4  + 45) = -30

Therefore F'(4) = -30 means that on day 4, the number of fish in the bay is decreasing at 30 fish per day.

(b) The absolute minimum is given as follows;

At a minimum or maximum value, F'(t) = 10·(3·t² - 24·t  + 45) = 0

Which gives;

3·(t - 5)·(t - 3) = 0

t = 5, or t = 3

At t = 5, we have;

f(5) = 10 × (5³ - 12 × 5² + 45 × 5 + 100) = 1,500

At t = 3, we have;

f(3) = 10 × (3³ - 12 × 3² + 45 × 3 + 100) = 1,540

Therefore, f(5) = 1,500 is a local maximum

However, at x = 0, we have;

f(0) = 10 × (0³ - 12 × 0² + 45 × 0 + 100) = 1000

At x = 8, we have;

f(8) = 10 × (8³ - 12 × 8² + 45 × 8 + 100) = 2,040

Therefore, the absolute minimum is given at t = 8, where f(t) = 1,000

(c) The values of t at which the rate of change in the number of fish in the

bay is decreasing is between the local maximum at t = 3, and the local

minimum at t = 5, which gives;

The rate of change of the number of fish is decreasing for values of t in the range;

• 3 ≤ t ≤ 5

(d)  P = 10·(t³ - 12·t² + 45·t + 100)

P' =  10·(3·t² - 24·t  + 45)

At time t = c, we have;

• P' =  10·(3·c² - 24·c  + 45)

Based on a similar question online, we have;

(c) The interval over which the rate of change is decreasing

(d) The rate of change of the number of pelicans flying near the bay at t = c

Learn more about differentiation of functions here:

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