Chapter 2.2, Problem 50E

a.

To determine

Find the engine speed that maximize torque, what is the maximum torque.

Answer to Problem 50E

Maximumtorque is 74.69 foot-pounds at 3.1 thousands of revolutions per minute.

Explanation of Solution

Given information: The engine torque y(in foot-pounds)of one modal of a car is given

by $y=-3.75x^2+23.2x+38.8$ ,where x is the speed (in thousands of revolutions per minute) of the engine.

Calculation:

  $y=-3.75x^2+23.2x+38.8$

Identify the coefficients

  $a=-3.75,b=23.2,c=\mathrm{38.8.}$

Since $a=-3.75<0$ , parabola opens down and function has a maximum value.

To find maximum value, calculate coordinates of the vertex.

  $\begin{array}{l}x=-b/2a=-23.2/2\times (-3.75)=-23.2/-7.5\approx \mathrm{3.1.}\\ $ f(3.1)=(-3.75)\times {(3.1)}^2+23.2\times (3.1)+38.8\approx 74.69$\end{array}$

Maximum torque is 74.69 foot-pounds at 3.1 thousands of revolutions per minute.

b.

To determine

What happens to the engine torque as the speed of the engine increases.

Answer to Problem 50E

At starts, torque increases as speed increases, but once that speed exceeds 3.1,torque decreases as speed increases.

Explanation of Solution

Given information: The engine torque y(in foot-pounds)of one modal of a car is given

by $y=-3.75x^2+23.2x+38.8$ ,where x is the speed (in thousands of revolutions per minute) of the engine.

Calculation:

  $y=-3.75x^2+23.2x+38.8$

Identify the coefficients

  $a=-3.75,b=23.2,c=\mathrm{38.8.}$

Since $a=-3.75<0$ , parabola opens down and function has a maximum value.

To find maximum value, calculate coordinates of the vertex.

  $\begin{array}{l}x=-b/2a=-23.2/2\times (-3.75)=-23.2/-7.5\approx \mathrm{3.1.}\\ $ f(3.1)=(-3.75)\times {(3.1)}^2+23.2\times (3.1)+38.8\approx 74.69$\end{array}$

The engine torque is increasing to the left of $x=3.1$ and decreasing to the right of $x=3.1$ .So,

At starts, torque increases as speed increases, but once that speed exceeds 3.1,torque decreases as speed increases.