Geometrically interpret the growth rate corresponding to the populatio

Question

Geometrically interpret the growth rate corresponding to the population growth model

$\frac{dx}{dt}=r_{t}x\left(\frac{x}{k_{0}}-1\right)\left(1-\frac{x}{k}\right),\quad 0<k_{0}<k$

$x(0)=x_{0},$

where r₁ , k, k are constants, k being the carrying capacity of the population x(t) . Hence find when the growth is maximum.

a) Consider the cubic total cost function

C = 0.004q³-0.8q³ +10q+5.

Assume that the price of q is 13 per unit. Find the output which yields maximum profit.

18 Feb 2025

Answer :

Word Count : 580

We are asked to solve two problems: one related to population growth and the other related to profit maximization. Let's solve both manually. --- ### 1. Population growth model The differential equation is: $$ \frac{dx}{dt} = r x \left(\frac{x}{k_0} - 1\right) \left(1 - \frac{x}{k}\right), \quad 0 < k_0 < k $$ with initial condition $x(0) = x_0$. Step 1: Geometrical interpretation * The growth rate $\frac{dx}{dt}$ is a cubic function of $x$. * It has three roots where $\frac{dx}{dt} = 0$: $$ x = 0, \quad x = k_0, \quad x = k $$ * The population grows if $\frac{dx}{dt} > 0$ and declines if $\frac{dx}{dt} < 0$. * Geometrically, the cubic curve passes through $(0,0), (k_0,0), (k,0)$. Between the roots, the slope determines growth: 1. $0 < x < k_0$ → $\frac{x}{k_0} - 1 < 0$, $1 - x/k > 0$ → $\frac{dx}{dt} < _____ _____ __________ _____ ___ ___ ____ ____ ________ ______.
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