Problem 28 - Contemporary Precalculus

Question

According to United Nations projections, the population of China (in millions) between now and 2050 is given by $$ y=-.00096 x^{3}-.1 x^{2}+11.3 x+1274 \quad(0 \leq x \leq 50) $$ where \(x=0\) corresponds to \(2000 .\) In what year will China reach its maximum population, and what will that population be?

Step-by-Step Solution

Answer

Answer: China will reach its maximum population of approximately 1438.84 million people in the year 2025.

Step 1: Find the derivative of the given function

We want to find the rate of change of the population with respect to the year x. To do this, we will take the derivative of y(x) with respect to x: $$ y'(x) = -0.00288x^2 -0.2x + 11.3 $$

Step 2: Set the derivative equal to zero and solve for x

We want to find the critical points of the function, which are the points where the derivative is equal to zero: $$ 0 = -0.00288x^2 -0.2x + 11.3 $$ Now, we will solve for x. This is a quadratic equation, so we can use the quadratic formula: $$ x = \frac{-(-0.2) \pm \sqrt{(-0.2)^2 - 4(-0.00288)(11.3)}}{2(-0.00288)} $$

Step 3: Evaluate x and identify maximum value

Now we will calculate the two possible x-values that satisfy the equation from step 2 and choose the one that is within the given domain (0 ≤ x ≤ 50): $$ x_1 = \frac{0.2 + \sqrt{0.04 + 0.13152}}{-0.00576} \approx 24.62, $$ $$ x_2 = \frac{0.2 - \sqrt{0.04 + 0.13152}}{-0.00576} \approx -17.29. $$ Since \(x_2\) is not within the domain, we discard it and consider only \(x_1 = 24.62\).

Step 4: Determine the year and maximum population

Now that we have found the critical point \(x_1 = 24.62\), we need to convert this x-value back to a year, and then use that to find the maximum population: $$ \text{year} = 2000 + x_1 \approx 2000 + 24.62 = 2024.62, $$ rounded to the nearest year, we get the year to be 2025. Now, we will find the population for this year: $$ y(24.62) \approx -0.00096(24.62)^3 - 0.1(24.62)^2 + 11.3(24.62) + 1274 \approx 1438.84. $$ So, the maximum population will be around 1438.84 million people. In conclusion, China will reach its maximum population of approximately 1438.84 million people in the year 2025.

Key Concepts

Population Modeling

Population modeling is a mathematical way to predict how a population will change over time. It uses equations to represent growth or decline. In this exercise, the population of China is modeled with a cubic equation. This equation includes terms like

• \(-0.00096x^3\)
• \(-0.1x^2\)
• \(11.3x\)
• constant term \(1274\)

Each term represents different forces that affect population changes, such as birth rates, death rates, and economic factors.
By analyzing this equation, we can estimate how the population will behave in the future. In this instance, this model aims to find the point at which China's population reaches its peak, before potentially decreasing thereafter.

Derivative

In mathematics, the derivative of a function represents its rate of change. For the given population model, taking the derivative helps us understand how the population is changing each year. The derivative of the function \(y(x)\) is calculated as \[y'(x) = -0.00288x^2 -0.2x + 11.3\]This expression shows how the growth rate changes with respect to \(x\), which is the number of years since 2000. The process involves differentiating each term of the original equation:

• The \(-0.00096x^3\) term becomes \(-0.00288x^2\) as its power is reduced by one.
• The \(-0.1x^2\) term becomes \(-0.2x\).
• The \(11.3x\) becomes a constant \(11.3\) since it’s a linear term.

No change occurs in constant terms. Finding where this derivative equals zero helps identify points where growth shifts from increasing to decreasing, which is vital in locating maximums or minimums.

Critical Points

Critical points occur where a function's derivative equals zero or is undefined. These are key for finding when a population reaches a maximum or minimum. Here, critical points are calculated by setting the derivative \(-0.00288x^2 - 0.2x + 11.3\) equal to zero. Solving this equation involves using the quadratic formula:\[x = \frac{-(-0.2) \pm \sqrt{(-0.2)^2 - 4(-0.00288)(11.3)}}{2(-0.00288)}\] This results in two solutions:

• \(x_1 = 24.62\)
• \(x_2 = -17.29\)

Since \(x\) represents years since 2000, and must be positive, \(x_1 = 24.62\) is the valid critical point. This point signals when maximum population occurs within the stipulated timeframe.

Quadratic Equation

A quadratic equation is one of the form \(ax^2 + bx + c = 0\). In this exercise, solving a quadratic equation is necessary for finding critical points. The relevant equation derived from the derivative is:\[-0.00288x^2 - 0.2x + 11.3 = 0\]Using the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], where \(a = -0.00288\), \(b = -0.2\), and \(c = 11.3\), allows calculating \(x\). The quadratic formula systematically handles each part:

• \(b^2 - 4ac\) checks the nature of the roots.
• Square root yields possible \(x\) values.

The exercise shows a real-life application, how solving quadratic equations help in analyzing population trends.