college algebra problem

Question

Flying too fast or slow wastes fuel. For a certain aircraft, miles per pound of fuel M was a function of air speed A in miles per hour, modeled by the formula

M = -0.000657A^2+0.127A-5.04. 

a .Use the formula to find the value of A that maximizes M. 

b. How many gallons of fuel were burned per hour if the aircraft was flown at 97 mph and got 1.10 mi/lb of fuel, which weighed 6.17 lb/gal? 

Answer 1

a.  The function for M is a quadratic equation (that means it is of the form:  y = ax^2 + bx + c).   The graph of a quadratic equation is a parabola,  and in this case,  since the leading coefficient (-.000657) is negative,  the parabola opens downward.   This means that the maximum value of M will be found at the 'vertex' of the parabola.  The vertex of any quadratic equation is located at:  -b/2a.   In this case,  -b/2a = -0.127 / 2(-0.000657) =  96.6 mi/hr. (approximately)

 

b.  In order to find the gallons of fuel burned per hour,  we simply have to do a few conversions:  Dividing 97 mi/hr by 1.1 mi/lb will give us:  88.1818...  lb/hr.  Now we just convert lbs to gallons by dividing by 6.17,  since there are 6.17 lbs for each gallon:  88.181818 lb/hr / 6.17 lb/gal = 14.292 gal/hr.

 

It is helpful to write out each step of the conversions in part b.  Remember that if you divide something like mi/hr by mi/lb,  that is the same as multiplying mi/hr by lb/mi.  This way you can see miles cancel out,  leaving you with lb/hr.  Then the same works for lb/hr divided by lb/gal to get gal/hr.