/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The population of a country is c... [FREE SOLUTION] | 91Ó°ÊÓ

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The population of a country is currently at 56 million and is forecast to rise by \(3.7 \%\) each year. It is capable of producing 2500 million units of food each year, and it is estimated that each member of the population requires a minimum of 65 units of food each year. At the moment, the extra food needed to satisfy this requirement is imported, but the government decides to increase food production at a constant rate each year, with the aim of making the country self-sufficient after 10 years. Find the annual rate of growth required to achieve this.

Short Answer

Expert verified
The annual growth rate required for food production to make the country self-sufficient in 10 years is approximately 7.7\text{%}.

Step by step solution

01

Determine the population in 10 years

The population increases by 3.7\text{%} each year. The population after 10 years can be calculated using the formula for exponential growth: \[ P_{10} = P_0 (1 + r)^n \]Where \( P_0 \) is the initial population, \( r \) is the growth rate, and \( n \) is the number of years.\[ P_{10} = 56 \text{ million} \times (1 + 0.037)^{10} \]Calculating this gives approximately:\[ P_{10} \approx 56 \times 1.437736 \approx 80.51216 \text{ million} \]
02

Calculate the food required in 10 years

Each person requires 65 units of food each year. Thus, the total food required after 10 years is:\[ F_{10} = P_{10} \times 65 \]Substituting the population found in Step 1:\[ F_{10} = 80.51216 \text{ million} \times 65 \]\[ F_{10} \approx 5233.29 \text{ million units} \]
03

Determine the required annual growth of food production

The country aims to produce this amount of food domestically after 10 years. The current food production is 2500 million units. Assume the food production grows at a constant rate \( g \). Thus, the formula for future value of food production is:\[ F_{10} = F_0 (1 + g)^{10} \]Where \( F_0 \) is the initial production of 2500 million units.So, we have:\[ 5233.29 = 2500(1 + g)^{10} \]Solving for \( g \) we get:\[ (1 + g)^{10} = \frac{5233.29}{2500} \]\[ (1 + g)^{10} = 2.093316 \]Taking the 10th root on both sides:\[ 1 + g = 2.093316^{\frac{1}{10}} \approx 1.077 \]\[ g \approx 0.077 \text{ or } 7.7\text{%} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

exponential growth formula
Exponential growth refers to the process by which a quantity increases over time at a rate proportional to its current value. This is a common concept in fields such as population studies and finance. The main formula used to calculate exponential growth is: \( P(t) = P_0 (1 + r)^t \). Here, \( P(t) \) is the value at time \( t \), \( P_0 \) is the initial value, \( r \) is the growth rate, and \( t \) is time. When we apply this to our given problem, with a population of 56 million growing at 3.7% per year, we want to find the population after 10 years. Using the formula, we substitute \( P_0 = 56 \text{ million} \), \( r = 0.037 \), and \( t = 10 \). This gives us: \[ P_{10} = 56 (1 + 0.037)^{10} \] Calculating this, we get: \[ P_{10} \approx 56 \times 1.437736 \approx 80.51216 \text{ million} \] This demonstrates how exponential growth can significantly increase a population over time.
future value computation
Future value computation is important for predicting how current values grow over time due to a set rate of increase. This applies to economies, personal finance, and resource management. The formula to compute the future value of a growing resource is similar to the exponential growth formula: \( FV = PV (1 + r)^n \). Where \( FV \) is the future value, \( PV \) is the present value, \( r \) is the growth rate, and \( n \) is the number of periods. Applying this to our problem, by substituting the initial food production as \( PV = 2500 \text{ million units} \) and solving for a future food requirement of \( 5233.29 \text{ million units} \), we rearrange the formula to solve for the growth rate \( g \): \[ 5233.29 = 2500 (1 + g)^{10} \] Taking the 10th root of both sides, we get: \[ (1 + g)^{10} = \frac{5233.29}{2500} \] This simplifies to: \[ 1 + g = 2.093316^{\frac{1}{10}} \approx 1.077 \] Thus, the annual growth rate needed is \( g \approx 7.7\text{\text{%%}} \).
resource allocation strategy
Creating a resource allocation strategy involves planning how resources like food, energy, and funds will be distributed over time to meet future needs. For our population and food example, such a strategy would include calculating the annual increase in food production necessary to support the growing population and become self-sufficient. Key steps include:
  • Determining future demand based on current growth rates
  • Calculating the required growth in production
  • Implementing policies to achieve the desired production increase
In our problem, the government needs to ensure that food production grows by 7.7% annually over the next 10 years. This strategic approach includes investments in agricultural technology, policies to encourage farming, and perhaps subsidies or financial supports to farmers to boost production.
annual growth rate calculation
Understanding annual growth rate calculation is crucial for predicting future values in various applications from business to environmental studies. The formula to calculate the future value informs us about the percentage by which a current value needs to increase yearly. The key equation here is: \[ (1 + g)^n = \frac{FV}{PV} \] Rearranging to solve for the annual growth rate \( g \), we get: \[ g = \bigg(\frac{FV}{PV}\bigg)^{\frac{1}{n}} - 1 \] For our food production problem, this means: \[ g = \bigg(\frac{5233.29}{2500}\bigg)^{\frac{1}{10}} - 1 \approx 1.077 - 1 \approx 0.077 \text{ or } 7.7\text{\text{%%}} \] Knowing this rate helps planners design effective policies and strategies to meet future demands efficiently.

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Most popular questions from this chapter

Find the future value of \(\$ 100\) compounded continuously at an annual rate of \(6 \%\) for 12 years.

A firm needs to choose between two projects, A and B. Project A involves an initial outlay of \(\$ 13500\) and yields \(\$ 18000\) in 2 years' time. Project \(B\) requires an outlay of \(\$ 9000\) and yields \(\$ 13000\) after 2 years. Which of these projects would you advise the firm to invest in if the annual market rate of interest is \(7 \%\) ?

If a principal, \(P\), is invested at \(r \%\) interest compounded annually then its future value, \(S\), after \(n\) years is given by $$ S=P\left(1+\frac{r}{100}\right)^{n} $$ (a) Use this formula to show that if an interest rate of \(r \%\) is compounded \(k\) times a year then after \(t\) years $$ S=P\left(1+\frac{r}{100 k}\right)^{k t} $$ (b) Show that if \(m=100 k / r\) then the formula in part (1) can be written as $$ S=P\left(\left(1+\frac{1}{m}\right)^{m}\right)^{r t / 100} $$ (c) Use the definition $$ \mathrm{e}=\lim _{m \rightarrow \infty}\left(1+\frac{1}{m}\right)^{m} $$ to deduce that if the interest is compounded with ever-increasing frequency (that is, continuously) then $$ S=P \mathrm{e}^{r / 100} $$

(a) Current monthly output from a factory is 25000 . In a recession, this is expected to fall by \(65 \%\). Estimate the new level of output. (b) As a result of a modernization programme, a firm is able to reduce the size of its workforce by \(24 \%\). If it now employs 570 workers, how many people did it employ before restructuring? (c) Shares originally worth \(\$ 10.50\) fall in a stock market crash to \(\$ 2.10\). Find the percentage decrease.

At the beginning of a month, a customer owes a credit card company \(\$ 8480\). In the middle of the month, the customer repays \(\$ A\), where \(A<\$ 8480\), and at the end of the month the company adds interest at a rate of \(6 \%\) of the outstanding debt. This process is repeated with the customer continuing to pay off the same amount, \(\$ A\), each month. (a) Find the value of \(A\) for which the customer still owes \(\$ 8480\) at the start of each month. (b) If \(A=1000\), calculate the amount owing at the end of the eighth month. (c) Show that the value of \(A\) for which the whole amount owing is exactly paid off after the \(n\)th payment is given by $$ A=\frac{8480 R^{n-1}(R-1)}{R^{n}-1} \text { where } R=1.06 $$ (d) Find the value of \(A\) if the debt is to be paid off exactly after 2 years.

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