College

The engine torque [tex]y[/tex] (in foot-pounds) of one model of a car is given by [tex]y=-3.75 x^2+23.2 x+38.8[/tex], where [tex]x[/tex] is the speed of the engine. Find the engine speed that maximizes the torque.

Answer :

To find the engine speed that maximizes the torque, we need to determine the vertex of the parabola represented by the quadratic equation [tex]\( y = -3.75x^2 + 23.2x + 38.8 \)[/tex].

This equation is in the standard form of a quadratic: [tex]\( y = ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -3.75 \)[/tex]
- [tex]\( b = 23.2 \)[/tex]
- [tex]\( c = 38.8 \)[/tex]

The vertex of a parabola given by a quadratic equation can be found using the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]

This formula helps us find the speed (x) at which the torque (y) is maximized, because the parabola opens downwards (since [tex]\( a \)[/tex] is negative).

Let's plug in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into this formula:
[tex]\[ x = -\frac{23.2}{2 \times (-3.75)} \][/tex]

Calculating the value:
[tex]\[ x = -\frac{23.2}{-7.5} \][/tex]

[tex]\[ x \approx 3.0933 \][/tex]

Therefore, the engine speed that maximizes the torque is approximately 3.09 (rounded to two decimal places). This is the optimal speed at which the engine torque is at its maximum.