Differential Equations 3 Question 12
12. A country has food deficit of $10 %$. Its population grows continuously at a rate of $3 %$ per year. Its annual food production every year is $4 %$ more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self- sufficient in food after $n$ years, where $n$ is the smallest integer bigger than or equal to $\frac{\ln 10-\ln 9}{\ln (1.04)-(0.03)}$.
$(2000,10 M)$
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Answer:
Correct Answer: 12. $\left(x^{2}+y^{2}=2 x\right)$ 14. $\log _{\frac{3}{4}} \frac{1}{2}$
Solution:
- Let $X _0$ be initial population of the country and $Y _0$ be its initial food production. Let the average consumption be $a$ unit. Therefore, food required initially $a X _0$. It is given
$$ Y _p=a X _0 \quad \frac{90}{100}=0.9 a X _0 $$
Let $X$ be the population of the country in year $t$.
Then, $\quad \frac{d X}{d t}=$ Rate of change of population
$$ \begin{array}{lc} & =\frac{3}{100} X=0.03 X \\ \Rightarrow & \frac{d X}{X}=0.03 d t \Rightarrow \quad \int \frac{d X}{X}=\int 0.03 d t \\ \Rightarrow & \log X=0.03 t+c \\ \Rightarrow \quad X=A \cdot e^{0.03 t}, \text { where } A=e^{c} \\ \text { At } & t=0, X=X _0, \text { thus } X _0=A \\ \therefore & X=X _0 e^{0.03 t} \end{array} $$
Let $Y$ be the food production in year $t$.
Then, $\quad Y=Y _0 \quad 1+\frac{4}{100}^{t}=0.9 a X _0(1.04)^{t}$
$\because \quad Y _0=0.9 a X _0$
Food consumption in the year $t$ is $a X _0 e^{0.03 t}$.
Again, $\quad Y-X \geq 0$
$$ \begin{array}{ll} \Rightarrow & 0.9 X _0 a(1.04)^{t}>a X _0 e^{0.03 t} \\ \Rightarrow & \frac{(1.04)^{t}}{e^{0.03 t}}>\frac{1}{0.9}=\frac{10}{9} . \end{array} $$
[given]
Taking log on both sides, we get
$$ t[\log (1.04)-0.03] \geq \log 10-\log 9 $$
$$ \Rightarrow \quad t \geq \frac{\log 10-\log 9}{\log (1.04)-0.03} $$
Thus, the least integral values of the year $n$, when the country becomes self-sufficient is the smallest integer greater than or equal to $\frac{\log 10-\log 9}{\log (1.04)-0.03}$.
