Tag Archives: Optimisation

March 25, 2025 · 7:32 am

Deriving the Logistic Growth Equation

The logistic differential equation

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

where r is the growth parameter and k is the carrying capacity.

And the maximum rate of increase happens when P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}

    \begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}
Hence,
\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}
1=A(k-P)+BP
When P=0,
1=Ak\Rightarrow A=\frac{1}{k}
When P=k,
1=BK\Rightarrow B=\frac{1}{k}

So our equation is,

    \begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}

    \begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}

When t=0, P=P_0,

    \begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}

The equation is now

    \begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}

    \begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}

    \begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}

Divide by e^{rkt}

    \begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

    \begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

Proving the Maximum Rate of Increase Happens When P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}

    \begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}rP(k-P)(k-2P)=0\end{equation}

Hence P=k or P=\frac{k}{2}

(1)   \begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}

Substitute P=k into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}

Hence, not a maximum.

Substitute P=\frac{k}{2} into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}

-2\frac{r^2k^4}{4}\le 0 For all values of P, r and k.

Hence maximum when P=\frac{k}{2}

We will look at a worked example in the next post.

Leave a Comment

Filed under Differential Equations, Differentiation, Implicit, Logistic Growth, Optimisation, Product Rule, Uncategorized, Year 12 Specialist Mathematics

Tagged as , , ,

September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

Leave a Comment

Filed under Differentiation, Optimisation, Year 12 Mathematical Methods

Tagged as , ,