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A model for the population p(t) in a suburb of a large city is given by the initial-value problem dp dt = p(10−1 − 10−7p), p(0) = 6000, where t is measured in months. what is the limiting value of the population? at what time will the population be equal to one-half of this limiting value? (round your answer to one decimal place.

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segunare233

Answer:

T = 200 months

Step-by-step explanation:

Given the model to be

dp/dt = p(10-1-10-7p) where p(0) = 6000

dp/dt = p(-1-7p)

dp/dt = -p+7p²

But since P(0) = 6000

We plug this value into the equation

P = -6000+7(6000)²

P = -6000+252000000

P = 251994000. .limiting value

Differentiating dp/dt w.r.t p

dp/dt = -1+14p

dp'/dt = 14p-1

Plugging p(0) = 6000 into the above we have that

dp'/dt = 14(6000)-1

P' = 84000

Time population would be equal to 3/2 the limiting value

T = 3/2× 84000 = 126000

T = 25199400/126000

T = 199.99months

T = 200 months.

zubam2002

Answer:

a) -29.4 ≤ p ≤ 29.2

b) 4 months

Step-by-step explanation:

dp/dt = p(10 - 1 - 10 - 7p) = p(- 1 - 7p) = -p - 7p²

p = ∫(-p - 7p²)dt = -tp - 7tp² ----------- (1)

For p(0) = 6000

Hence, dp/dt = 0

∴ -p - 7p² = 6000

7p² + p + 6000 = 0

Using the formular, p = -b±√b² - 4ac/2a, where, a = 7; b = 1; c = 6000

p = -1±√1 - 4(7)(6000)/2*7 = -1±√-167999/14 =

p = -1±(- 409.88)/14= -410.88/14 or 408.88/14 = -29.4 or 29.2

∴ The limiting value of the population:

-29.4≤ p ≤ 29.2

b) When p = -29.4/2  or 29.2/2 = ±15

From (1), we have it that, 7tp² + tp = 6000

By substituting the value, p = -15

7t(-15)² - 15t = 6000

1575t - 15t = 6000

∴ t = 6000/1590 = 3.77

It wiil take close to 4 months for the population to be equal to one - haif of the limiting value