M(x)=-0.015x^2+1.31x-7.3, [30,60], represent the miles per gallon obtained at a speed of x mph. Find the absolute maximum miles per gallon and absolute minimum and the speeds at which they occur.

Question

Need help with this question for my calculus homework.

Answer 1

To find the extrema, differentiate M: -0.03x+1.31=0. Solve for x=1.31/0.03=131/3=43.67 approx.

M(131/3)=21.30 mpg approx. M(30)=18.5 mpg, M(60)=17.3 mpg. So in the given range the maximum mpg is 21.30 mpg at speed 43.67 mph and the minimum is 17.3 mpg at 60 mph.