Section 2.4 question: 2 Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month. If the yard can only sustain 140 plants, use a logistic growth model to estimate the number of plants after 5 months. plants

Section 24 question 2 Suppose 27 blackberry plants started growing in a yard Absent constraint the blackberry plants will spread by 80 a month If the yard can o class=

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The question indicates that we should use a logistic model.

This can be given as

[tex]P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_o}{P_0}[/tex]

In this case. K is the carrying capacity =140

Po is the initial population=27

[tex]\begin{gathered} \therefore A=\frac{140-27}{27} \\ A=4.185 \end{gathered}[/tex]

This then brings the model to be

[tex]P(t)=\frac{140}{1+4.185e^{-kt}}[/tex]

The next step would be to find the value of k

Since the blackberry plants increase by 80% every month. Therefore, for 1 month we would have

[tex]27+\frac{80}{100}\times27=48.6[/tex]

This implies that for that first month we would have

[tex]\begin{gathered} \frac{140}{1+4.185e^{-k\times1}}=48.6 \\ \frac{140}{1+4.185e^{-k}}=48.6 \\ 140=48.6+203.391e^{-k} \\ 91.4=203.391e^{-k} \\ e^{-k}=\frac{91.4}{203.391} \\ e^{-k}=0.44938 \\ \ln (e^{-k})=\ln 0.44938 \\ -k=-0.7999 \\ k=0.7999 \end{gathered}[/tex]

Therefore for 5 months, we would have

[tex]\begin{gathered} P(5)=\frac{140}{1+4.185e^{-0.7999\times5}} \\ P(5)=\frac{140}{1+4.185e^{-3.9995}} \\ P(5)=\frac{140}{1.0770352} \\ P(5)=129.986\approx130 \end{gathered}[/tex]

Answer: Using the logistic model the estimated number of plants after 5 months becomes

[tex]130[/tex]

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