2. (5 pts) the number of us smokers each year can be modeled by the equation s(x)= - 0.03x² + 5x - 173.2…

2. (5 pts) the number of us smokers each year can be modeled by the equation s(x)= - 0.03x² + 5x - 173.2 million smokers x years after 1900, for 60 ≤ x ≤ 100. checkpoint: s(1)= - 168.23 a. write the equation you would need to solve to find the x - values of the critical points of s(x) without the aid of your calculator. - 0.06x + 5 = 0 b. find the absolute extremes on the interval 60,100. give both the input and output values. absolute maximum on 60,100: absolute minimum on 60,100: c. interpret your answers from part b by completing the following sentences. between 1960 and 2000, the number of us smokers was highest in _, when there were _ smokers. between 1960 and 2000, the number of us smokers was lowest in _, when there were _ smokers. 3. (5 pts) prelkers’ patties is a food truck which serves high quality, artisan hamburgers. the amount of profit the truck makes in a day depends on the number of burgers it sells, with data values given in the table below. number of burgers sold 20 40 60 80 100 120 profit (dollars) - 182.10 59.59 181.36 269.19 262.13 186.34 a. find a quadratic model for the given data. p(b)= - 0.1116² + 16.652b - 347.345 gives where b is _, _ b. use your calculator and the unrounded model to find the relative maximum of the function p(b). state the maximum as a point with three decimal places: 867.186 complete the sentence of interpretation with rounding in context: to maximize profit, prelkers’ patties should sell _ in a day, which would yield a profit of _.

2. (5 pts) the number of us smokers each year can be modeled by the equation s(x)= - 0.03x² + 5x - 173.2 million smokers x years after 1900, for 60 ≤ x ≤ 100. checkpoint: s(1)= - 168.23 a. write the equation you would need to solve to find the x - values of the critical points of s(x) without the aid of your calculator. - 0.06x + 5 = 0 b. find the absolute extremes on the interval 60,100. give both the input and output values. absolute maximum on 60,100: absolute minimum on 60,100: c. interpret your answers from part b by completing the following sentences. between 1960 and 2000, the number of us smokers was highest in _, when there were _ smokers. between 1960 and 2000, the number of us smokers was lowest in _, when there were _ smokers. 3. (5 pts) prelkers’ patties is a food truck which serves high quality, artisan hamburgers. the amount of profit the truck makes in a day depends on the number of burgers it sells, with data values given in the table below. number of burgers sold 20 40 60 80 100 120 profit (dollars) - 182.10 59.59 181.36 269.19 262.13 186.34 a. find a quadratic model for the given data. p(b)= - 0.1116² + 16.652b - 347.345 gives where b is _, _ b. use your calculator and the unrounded model to find the relative maximum of the function p(b). state the maximum as a point with three decimal places: 867.186 complete the sentence of interpretation with rounding in context: to maximize profit, prelkers’ patties should sell _ in a day, which would yield a profit of _.

Answer

Explanation:

Step1: Recall critical - point formula

For a function $y = f(x)$, critical points are found by setting $f^\prime(x)=0$. Given $S(x)=- 0.03x^{2}+5x - 173.2$, its derivative $S^\prime(x)=-0.06x + 5$. So the equation to find critical points is $-0.06x + 5=0$.

Step2: Solve for critical - point $x$

Solve $-0.06x + 5 = 0$ for $x$. First, subtract 5 from both sides: $-0.06x=-5$. Then divide both sides by $-0.06$: $x=\frac{5}{0.06}=\frac{500}{6}=\frac{250}{3}\approx83.33$.

Step3: Evaluate $S(x)$ at critical - point and endpoints

Evaluate $S(x)$ at $x = 60$, $x=\frac{250}{3}$, and $x = 100$. $S(60)=-0.03\times60^{2}+5\times60 - 173.2=-0.03\times3600 + 300-173.2=-108 + 300-173.2=18.8$. $S(\frac{250}{3})=-0.03\times(\frac{250}{3})^{2}+5\times\frac{250}{3}-173.2=-0.03\times\frac{62500}{9}+\frac{1250}{3}-173.2=-\frac{1875}{9}+\frac{3750}{9}-173.2=\frac{1875}{9}-173.2\approx208.33 - 173.2 = 35.13$. $S(100)=-0.03\times100^{2}+5\times100 - 173.2=-300 + 500-173.2=26.8$. So the absolute maximum on $[60,100]$ is at $x=\frac{250}{3}\approx83.33$ and $S(\frac{250}{3})\approx35.13$ (in million smokers), and the absolute minimum on $[60,100]$ is at $x = 60$ and $S(60)=18.8$ (in million smokers). For the profit - function part:

Step1: Quadratic model form

A quadratic model is of the form $P(b)=ab^{2}+bb + c$. Using regression (not shown here as the model is already given $P(b)=-0.111b^{2}+16.652b - 347.345$), where $b$ is the number of burgers sold.

Step2: Find vertex of quadratic function

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex (which gives the maximum for $a<0$) is $x =-\frac{b}{2a}$. Here $a=-0.111$ and $b = 16.652$. So $b=-\frac{16.652}{2\times(-0.111)}=\frac{16.652}{0.222}\approx74.92$. $P(74.92)=-0.111\times(74.92)^{2}+16.652\times74.92-347.345\approx301.10$.

Answer:

a. The equation to find the $x$-values of the critical points of $S(x)$ is $-0.06x + 5 = 0$. b. Absolute maximum on $[60,100]$: Input $x\approx83.33$, Output $S(x)\approx35.13$ (million smokers). Absolute minimum on $[60,100]$: Input $x = 60$, Output $S(x)=18.8$ (million smokers). c. Between 1960 and 2000, the number of US smokers was highest in 1983 (since $x$ is years after 1900 and $x\approx83.33$), when there were 35.13 million smokers. Between 1960 and 2000, the number of US smokers was lowest in 1960, when there were 18.8 million smokers. 3. a. $P(b)=-0.111b^{2}+16.652b - 347.345$ where $b$ is the number of burgers sold. b. To maximize profit, Prelkers’ Patties should sell 74.92 in a day, which would yield a profit of $$301.10$.