opatovaL

2021-03-07

Population The resident population P (in millions) of the United States from 2000 through 2013 can be modeled by $P=-0.00232{t}^{3}+0.0151{y}^{2}+2.83t+281.8,0\le t\le 13,$ where $t=0$ corresponds to 2000.
(Source: U.S. Census Bureau)
Make a conjecture about the maximum and minimum populations of the United States from 2000 to 2013.
Analytically find the maximum and minimum populations over the interval.
The brief paragrah while comparing a conjecture with the minimum population was 281.8 million in 2000 and the maximum population was 316.1 million in 2013.
We need to calculate: The absolute extrema of the popullation $P=-0.00232{t}^{3}+0.0151{y}^{2}+2.83t+281.8,0\le t\le 13$ over the closed interval [0, 13].

brawnyN

Used formula:
The derivative of power function given by
$\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$
Procedure to find the extrema of the continuous function f on closed interval [a, b].
Step 1: Find the derivative of the function f.
Step 2: Find the critical points of f in the open interval (a, b).
Step 3: Determine the value of f at each of the critical numbers in the open interval (a,b).
Step 4: Determine the value of f at each of the end-points a and b.
Step 5: The least of these values is the minimum and the greatest is the maximum.
Calculation:
Consider the function $P=P=-0.00232{t}^{3}+0.0151{y}^{2}+2.83t+281.8,$
Step 1. Determine first derivative of the function $P=P=-0.00232{t}^{3}+0.0151{y}^{2}+2.83t+281.8,0$
$P‘=-0.0069{t}^{2}+0.03t+2.83$
Step 2. Find out the critical points of P. It occurs when $P‘=0$ or P` underfined.
$P‘=0$
$-0.0069{t}^{2}+0.03t+2.83=0$
$t=\frac{-0.03±{\sqrt{0.03}}^{2}-4\left(-0.0069\right)\left(2.83\right)}{2\left(-0.0069\right)}$
$=\frac{-0.03±\sqrt{0.0009+0.078108}}{-0.0138}$
$=\frac{-0.03±\sqrt{0.079008}}{-0.0138}$
$=\frac{-0.03±0.281}{-0.0138}$
Further solving,
$t=\frac{-0.03±0.281}{-0.0138}$
$=-18.19$
And,
$t=\frac{-0.03-0.281}{-0.0138}$
$=22.53$
This gives,
$t=-18.19,22.53$
For end-point $t=0,$
Substitute $t=0$ in the function $P=-0.00232{t}^{3}+0.0151{y}^{2}+2.83t+281.8,$
$P=-0.0023\left(0{\right)}^{3}+0.015\left(0{\right)}^{2}+2.83\left(0\right)+281.8,$
$=281.8$
For end-point $t=13,$
Substitute $t=13$ in the function $P=-{0.0023}^{3}+{0.015}^{2}+2.83t+281.8,$
$P=-0.0023\left(13{\right)}^{3}+0.015\left(13{\right)}^{2}+2.83\left(13\right)+281.8,$
$=-0.0023\left(2197\right)+0.015\left(169\right)+3.83\left(13\right)+281.8$
$=-5.531+2.535+36.79+281.8$
$=316.1$
Use all the information to form the table as,
$\begin{array}{|ccc|}\hline t-value& s=0& r=13\\ P& 281.8& 316.1\\ Conclusion& Minimum& Maximum\\ \hline\end{array}$
From the table, it can be concluded that the minimum population was 281.8 million in 2000 and the maximum population was 316.1 million in 2013.

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