a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as
... c = k·s² + m·s + n
Filling in the given values gives three equations in k, m, and n.
[tex]28 = k\cdot 10^2+m\cdot 10+n\\21=k\cdot 20^2+m\cdot 20+n\\16=k\cdot 30^2+m\cdot 30+n[/tex]
Subtracting each equation from the one after gives
[tex]-7=300k+10m\\-5=500k+10m[/tex]
Subtracting the first of these equations from the second gives
[tex]2=200k\\\\k=\dfrac{2}{200}=0.01[/tex]
Using the next previous equation, we can find m.
[tex]-5=500\cdot 0.01+10m\\\\m=\dfrac{-10}{10}=-1[/tex]
Then from the first equation
[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]
There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with
... c = 0.01s² - s + 37
b) At 150 kph, the cost is predicted to be
... c = 0.01·150² -150 +37 = 112 . . . cents/km
c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.
d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.
