Question

The fuel efficiency for a certain midsize car is given by E(v) = −0.015v2 + 1.350v + 3.4 where E(v) is the fuel efficiency in miles per gallon for a car traveling v miles per hour. What speed will yield the maximum fuel efficiency? Round to the nearest mile per hour.

Answer 1

The speed that yields the maximum fuel efficiency for the car described by the equation E(v) = −0.015v^2 + 1.350v + 3.4 is 45 miles per hour, which is the vertex of the downward-opening parabola.

Step-by-step explanation:

To find the speed that will yield the maximum fuel efficiency for the given car, you need to analyze the quadratic equation E(v) = −0.015v2 + 1.350v + 3.4. This is a parabola that opens downwards, since the coefficient of the v2 term is negative. The vertex of the parabola represents the maximum point, and thus the maximum fuel efficiency. To find the vertex, you can use the formula -b/(2a) from the standard quadratic form ax2 + bx + c.

Applying the formula to the equation E(v) gives us v = −1.350 / (2 * (−0.015)) = 45 miles per hour. Therefore, the speed that will yield the maximum fuel efficiency is 45 mph, which is the vertex of the parabola represented by the given equation.