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Alina [70]
3 years ago
15

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

W the number of wolves. For example, if you found three equilibrium solutions, one with 100 rabbits and 10 wolves, one with 200 rabbits and 20 wolves, and one with 300 rabbits and 30 wolves, you would enter (100,10),(200,20),(300,30). Do not round fractional answers to the nearest integer.
Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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Answer:

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X/100% x 140 = 50.4

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Complete question :

A population of 30 deer are introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain 600 deer. The population would grow by 30 percent per year

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Answer:

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Step-by-step explanation:

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