We have to find at what speed will the car get the maximum mileage.
This can be done calculating the first derivative for m(x) in respect to x:
[tex]\begin{gathered} m(x)=-0.04x^2+3.6x-49 \\ \frac{dm}{dx}=-0.04(2x)+3.6(1)-49(0)=-0.08x+3.6 \end{gathered}[/tex]
Now we equal it to 0 and solve for x:
[tex]\begin{gathered} \frac{dm}{dx}=0 \\ -0.08x+3.6=0 \\ 3.6=0.08x \\ x=\frac{3.6}{0.08} \\ x=45 \end{gathered}[/tex]
As x = 45 is within the valid interval (between 20 and 70), we know it is a valid solution.
The maximum mileage happens at 45 miles per hour.
We can now calculate this gas mileage as m(45):
[tex]\begin{gathered} m(45)=-0.04(45)^2+3.6(45)-49 \\ m(45)=-0.04\cdot2025+162-49 \\ m(45)=-81+162-49 \\ m(45)=32 \end{gathered}[/tex]
The maximum mileage is 32 miles per gallon.
We can see it in a graph as:
Answer: the maximum mileage happens at a speed of 45 miles per hour, and the maximum mileage is 32 miles per gallon.
