Population The resident population P (in millions) of the United States from 2000 through 2013 can be modeled by P = -0.00232t^{3} + 0.0151y^{2} + 2.8

Used formula:
The derivative of power function given by
$\frac{d}{dx}(x^n)=n{x}^{n-1}$
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 $P=P=-0.00232t^3+0.0151y^2+2.83t+281.8,$
Step 1. Determine first derivative of the function $P=P=-0.00232t^3+0.0151y^2+2.83t+281.8,0$
$P‘=-0.0069t^2+0.03t+2.83$
Step 2. Find out the critical points of P. It occurs when $P‘=0$ or P` underfined.
$P‘=0$
$-0.0069t^2+0.03t+2.83=0$
Apply quadratix formula as,
$t=\frac{-0.03\pm {\sqrt{0.03}}^2-4(-0.0069)(2.83)}{2(-0.0069)}$
$=\frac{-0.03\pm \sqrt{0.0009+0.078108}}{-0.0138}$
$=\frac{-0.03\pm \sqrt{0.079008}}{-0.0138}$
$=\frac{-0.03\pm 0.281}{-0.0138}$
Further solving,
$t=\frac{-0.03\pm 0.281}{-0.0138}$
$=-18.19$
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 $P=-0.00232t^3+0.0151y^2+2.83t+281.8,$
$P=-0.0023(0{)}^3+0.015(0{)}^2+2.83(0)+281.8,$
$=281.8$
For end-point $t=13,$
Substitute $t=13$ in the function $P=-{0.0023}^3+{0.015}^2+2.83t+281.8,$
$P=-0.0023(13{)}^3+0.015(13{)}^2+2.83(13)+281.8,$
$=-0.0023(2197)+0.015(169)+3.83(13)+281.8$
$=-5.531+2.535+36.79+281.8$
$=316.1$
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.