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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.

Sagot :

MrRoyal

Answer:

[tex](0,0)[/tex]   [tex](4000,0)[/tex] and [tex](500,79)[/tex]

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

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

[tex]\frac{dW}{dt} = -0.02W + 0.00004RW[/tex]

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

So, we have:

[tex]0.09R(1 - 0.00025R) - 0.001RW = 0[/tex]

[tex]-0.02W + 0.00004RW = 0[/tex]

Factor out R in [tex]0.09R(1 - 0.00025R) - 0.001RW = 0[/tex]

[tex]R(0.09(1 - 0.00025R) - 0.001W) = 0[/tex]

Split

[tex]R = 0[/tex]   or [tex]0.09(1 - 0.00025R) - 0.001W = 0[/tex]

[tex]R = 0[/tex]   or  [tex]0.09 - 2.25 * 10^{-5}R - 0.001W = 0[/tex]

Factor out W in [tex]-0.02W + 0.00004RW = 0[/tex]

[tex]W(-0.02 + 0.00004R) = 0[/tex]

Split

[tex]W = 0[/tex] or [tex]-0.02 + 0.00004R = 0[/tex]

Solve for R

[tex]-0.02 + 0.00004R = 0[/tex]

[tex]0.00004R = 0.02[/tex]

Make R the subject

[tex]R = \frac{0.02}{0.00004}[/tex]

[tex]R = 500[/tex]

When [tex]R = 500[/tex], we have:

[tex]0.09 - 2.25 * 10^{-5}R - 0.001W = 0[/tex]

[tex]0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0[/tex]

[tex]0.09 -0.01125 - 0.001W = 0[/tex]

[tex]0.07875 - 0.001W = 0[/tex]

Collect like terms

[tex]- 0.001W = -0.07875[/tex]

Solve for W

[tex]W = \frac{-0.07875}{ - 0.001}[/tex]

[tex]W = 78.75[/tex]

[tex]W \approx 79[/tex]

[tex](R,W) \to (500,79)[/tex]

When [tex]W = 0[/tex], we have:

[tex]0.09 - 2.25 * 10^{-5}R - 0.001W = 0[/tex]

[tex]0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0[/tex]

[tex]0.09 - 2.25 * 10^{-5}R = 0[/tex]

Collect like terms

[tex]- 2.25 * 10^{-5}R = -0.09[/tex]

Solve for R

[tex]R = \frac{-0.09}{- 2.25 * 10^{-5}}[/tex]

[tex]R = 4000[/tex]

So, we have:

[tex](R,W) \to (4000,0)[/tex]

When [tex]R =0[/tex], we have:

[tex]-0.02W + 0.00004RW = 0[/tex]

[tex]-0.02W + 0.00004W*0 = 0[/tex]

[tex]-0.02W + 0 = 0[/tex]

[tex]-0.02W = 0[/tex]

[tex]W=0[/tex]

So, we have:

[tex](R,W) \to (0,0)[/tex]

Hence, the points of equilibrium are:

[tex](0,0)[/tex]   [tex](4000,0)[/tex] and [tex](500,79)[/tex]

View image MrRoyal
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