Respuesta :
Answer:
Option B - 6.6 miles per second
Step-by-step explanation:
Given : The speed of a runner, in miles per hour, on a straight trail is modeled by [tex]\frac{1}{10}(-2m^3+9m^2-12m)+7[/tex], where m is the runner's distance, in miles, from the start of the trail.
To find : What is the maximum speed of the runner for [tex]0\leq m\leq 3[/tex]?
Solution :
The model is given by,
[tex]f(m)=\frac{1}{10}(-2m^3+9m^2-12m)+7[/tex]
[tex]f(m)=\frac{-2}{10}m^3+\frac{9}{10}m^2-\frac{12}{10}m+7[/tex]
Derivate w.r.t m,
[tex]f'(m)=\frac{-2\times 3}{10}m^2+\frac{9\times 2}{10}m-\frac{12}{10}[/tex]
[tex]f'(m)=\frac{-3}{5}m^2+\frac{9}{5}m-\frac{6}{5}[/tex]
[tex]f'(m)=\frac{-3m^2+9m-6}{5}[/tex]
[tex]f'(m)=\frac{-3}{5}(m^2-3m+2)[/tex]
[tex]f'(m)=\frac{-3}{5}(m-1)(m-2)[/tex]
For critical point put it to zero,
[tex]\frac{-3}{5}(m-1)(m-2)=0[/tex]
[tex](m-1)(m-2)=0[/tex]
[tex]m=1,2[/tex]
Derivate again w.r.t m,
[tex]f''(m)=-\frac{-3}{5}(2m-3)[/tex]
At m=1,
[tex]f''(m)=\frac{-3}{5}(2(1)-3)=\frac{3}{5}>0[/tex]
It is minimum at m=1.
At m=2,
[tex]f''(m)=\frac{-3}{5}(2(2)-3)=\frac{-3}{5}<0[/tex]
It is maximum at m=2.
So, the maximum speed of the runner for [tex]0\leq m\leq 3[/tex] is given by
[tex]f(2)=\frac{1}{10}(-2(2)^3+9(2)^2-12(2))+7[/tex]
[tex]f(2)=\frac{1}{10}(-16+36-24)+7[/tex]
[tex]f(2)=\frac{1}{10}(-4)+7[/tex]
[tex]f(2)=6.6[/tex]
Therefore, the maximum speed of the runner for [tex]0\leq m\leq 3[/tex] is 6.6 miles per second.
Option B is correct.
In given problem, first we have to find the frontier of interval and then calculate the critical point of the function. After comparing all the value of function, the smallest value is absolute minimum and the largest value is absolute maximum. Hence, the maximum speed of the runner is 7 miles per second and the correct option is B.
Given:
The speed of a runner, in miles per hour, on a straight trail is modeled by [tex]\frac{1}{10}(-2m^3+9m^2-12m) +7[/tex] where m is the runner's distance, in miles, from the start of the trail.
Find the function at frontier of interval.
[tex]f(0)=\frac{1}{10}(-2\times 0^3+9 \times 0^2-12\times 0) +7=7\\f(3)=\frac{-2}{10}\times 3^3+\frac{9}{10}\times m^2-\frac{12}{10}\times 3 +7=\frac{61}{10}[/tex]
To find the critical point of the given function, find derivative of the function.
[tex]f'(m)=\frac{-3}{5}m^2+\frac{9}{5}m-\frac{6}{5}\\f'(m)=\frac{-3}{5}(m-1)(m-2)[/tex]
Putting the value of [tex]f'(m)=0[/tex]
[tex]\frac{-3}{5}(m-1)(m-2)=0\\m=1,\;2[/tex]
Find the function at critical point [tex]x_1=1[/tex]
[tex]f(0)=\frac{1}{10}(-2\times 1^3+9 \times 1^2-12\times 1) +7=\frac{13}{2}[/tex]
Find the function at critical point [tex]x_1=2[/tex]
[tex]f(0)=\frac{1}{10}(-2\times 2^3+9 \times 2^2-12\times 2) +7=\frac{33}{5}[/tex]
Compare the value of function at all point.
[tex]f(0)>f(2)>f(1)>f(3)\rightarrow7>\frac{33}{5}>\frac{13}{2}>\frac{61}{10}[/tex]
From above analysis, runner attained maximum speed at m=0.
Therefore, the maximum speed of the runner is 7 miles per second.
Learn more about Application of derivatives here:
https://brainly.com/question/14101278
